a) tính thường

b) \(\left(x-1\right)\left(x+2\right)< 0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1>0\\x+2< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -2\end{cases}}\Leftrightarrow1< x< -2\left(ktm\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-1< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 1\\x>-2\end{cases}}\Leftrightarrow-2< x< 1\left(tm\right)\)

vậy

c)\(\left(x+\frac{3}{5}\right)\left(x+1\right)< 0\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{5}< 0\\x+1>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< -\frac{3}{5}\\x>-1\end{cases}}\Leftrightarrow-1< x< -\frac{3}{5}\left(tm\right)\)

d) \(\left(x-\frac{1}{3}\right)\left(x+\frac{2}{5}\right)>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>\frac{1}{3}\\x>-\frac{2}{5}\end{cases}}\Leftrightarrow x>\frac{1}{3}\left(tm\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< \frac{1}{3}\\x< -\frac{2}{5}\end{cases}}\Leftrightarrow x< \frac{-2}{5}\left(tm\right)\)

vậy ...

a) 5/2 - x + 4/5 = 2/3 + 4/7

<=> 33/10 - x = 26/21

<=> x = 433/210

b) ( x - 1 )( x + 2 ) < 0 ( cái " x " kia là nhân à :v )

Xét 2 trường hợp

1.\(\hept{\begin{cases}x-1>0\\x+2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>1\\x< -2\end{cases}}\)( loại )

2. \(\hept{\begin{cases}x-1< 0\\x+2>0\end{cases}}\Rightarrow\hept{\begin{cases}x< 1\\x>-2\end{cases}}\Rightarrow-2< x< 1\)

Vậy -2 < x < 1

c) ( x + 3/5 )( x + 1 ) < 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}x+\frac{3}{5}< 0\\x+1>0\end{cases}}\Rightarrow\hept{\begin{cases}x< -\frac{3}{5}\\x>-1\end{cases}}\Rightarrow-1< x< -\frac{3}{5}\)

2. \(\hept{\begin{cases}x+\frac{3}{5}>0\\x+1< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-\frac{3}{5}\\x< -1\end{cases}}\)( loại )

Vậy -1 < x < -3/5

d) ( x - 1/3 )( x + 2/5 ) > 0

Xét hai trường hợp :

1.\(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\Rightarrow\hept{\begin{cases}x>\frac{1}{3}\\x>-\frac{2}{5}\end{cases}}\Rightarrow x>\frac{1}{3}\)

2.\(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< \frac{1}{3}\\x< -\frac{2}{5}\end{cases}\Rightarrow}x< -\frac{2}{5}\)

Vây \(\orbr{\begin{cases}x>\frac{1}{3}\\x< -\frac{2}{5}\end{cases}}\)

Bài làm

a) 2( x + 1 ) - 4x  = 6

=> 2x + 2 - 4x     = 6

=> ( 2x - 4x ) + 2 = 6

=>       -2x     + 2 = 6

=>       -2x            = 4

=>          x             = -2

Vậy x = -2

b) 3( 2 - x ) + 4( 5 - x )     = 4

=> 6 - 3x + 20 - 4x           = 4

=> ( 6 +20 ) + ( -3x - 4x ) = 4

=>       26     -       7x        = 4

=>                         7x       = 22

=>                           x       = 22/7

Vậy x = 22/7

c) Cũng phân tích như hai câu trên rồi rút gọn ra, sử dụng tính chất phân phối đó, do là phân số nên mik k muốn làm.

d) ( x + 1 )( x - 3 ) = 0

=> \(\hept{\begin{cases}x+1=0\Rightarrow x=-1\\x-3=0\Rightarrow x=3\end{cases}}\)

Vậy x = -1; x = 3

# Học tốt #

Tìm x biết : 

a) \(2\left(x+1\right)-4x=6\)

\(\Rightarrow2x+2-4x=6\)

\(\Rightarrow2x-4x=6-2\)

\(\Rightarrow-2x=4\)

\(\Rightarrow x=-2\)

b) \(3\left(2-x\right)+4\left(5-x\right)=4\)

\(\Rightarrow6-3x+20-4x=4\)

\(\Rightarrow-3x-4x=4-6-20\)

\(\Rightarrow-7x=22\)

\(\Rightarrow x=-\frac{22}{7}\)

c) \(\frac{7}{3}.\left(x-\frac{4}{3}\right)+\frac{2}{5}.\left(4-\frac{1}{3}x\right)=0\)

\(\Rightarrow\frac{7}{3}x-\frac{28}{9}+\frac{8}{5}-\frac{2}{15}x=0\)

\(\Rightarrow\left(\frac{7}{3}x-\frac{2}{15}x\right)-\left(\frac{28}{9}-\frac{8}{5}\right)=0\)

\(\Rightarrow\frac{33}{15}x-\frac{68}{45}=0\)

\(\Rightarrow\frac{33}{15}.x=\frac{68}{45}\)

\(\Rightarrow x=\frac{68}{45}:\frac{33}{15}\)

\(\Rightarrow x=\frac{68}{99}\)

d) \(\left(x+1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

a. \(5.\left(x-2\right)+3.\left(x-2\right)=0\)

\(\Rightarrow8.\left(x-2\right)=0\)

\(\Rightarrow x-2=0:8\)

\(\Rightarrow x-2=0\)

\(\Rightarrow x=2\)

Vậy...

b. \(\dfrac{2}{3}+\dfrac{5}{2}:x=\dfrac{2}{4}\)

\(\Rightarrow\dfrac{5}{2}:x=\dfrac{2}{4}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{5}{2}:x=\dfrac{-1}{6}\)

\(\Rightarrow x=\dfrac{5}{2}:\dfrac{-1}{6}=-15\)

Vậy...

c. \(2.\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow x-\dfrac{1}{7}=0:2\)

\(\Rightarrow x-\dfrac{1}{7}=0\)

\(\Rightarrow x=\dfrac{1}{7}\)

Vậy...

d. \(\dfrac{11}{20}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}:\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{-3}{20}\)

Vậy...

e. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)

Vậy...

g. \(\dfrac{2}{3}x+\dfrac{5}{7}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}\)

\(\Rightarrow\dfrac{2}{3}x=\dfrac{-29}{70}\)

\(\Rightarrow x=\dfrac{-29}{70}:\dfrac{2}{3}=\dfrac{-87}{140}\)

Vậy...

a) \(|x+\frac{3}{4}|+|y-\frac{1}{5}|+|x+y+z|=0\)

\(\Rightarrow|x+\frac{3}{4}|=|y-\frac{1}{5}|=|x+y+z|=0\)

\(\Rightarrow|x+\frac{3}{4}|=0\)                           \(\Rightarrow|y-\frac{1}{5}|=0\)                                \(\Rightarrow|x+y+z|=0\)

\(\Rightarrow x+\frac{3}{4}=0\)                              \(\Rightarrow y-\frac{1}{5}=0\)                                      \(\Rightarrow x+y+z=0\)

\(x=\frac{-3}{4}\)                                                \(y=\frac{1}{5}\)                                                 thay x=-3/4; y=1/5 vào biểu thức trên

                                                                                                                                          ta có \(\frac{-3}{4}+\frac{1}{5}+z=0\)

                                                                                                                                                        \(z=0-\frac{-3}{4}-\frac{1}{5}\)

      VẬY X=-3/4; Y=1/5; Z=11/20

B) \(|3x-4|+\left|3y-5\right|=0\)

\(\Rightarrow\left|3x-4\right|=\left|3y-5\right|=0\)

\(\Rightarrow\left|3x-4\right|=0\)                                    \(\Rightarrow\left|3y-5\right|=0\)

\(3x-4=0\)                                                    \(3y-5=0\)

\(3x=4\)                                                                    \(3y=5\)
\(x=\frac{4}{3}\)                                                                       \(y=\frac{5}{3}\)

VẬY X= 4/3; Y=5/3

C) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)

ĐỂ \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|;\left|y-\frac{2}{5}\right|;\left|z+\frac{1}{2}\right|< 0\)

MÀ GIÁ TRỊ TUYỆT ĐỐI LUÔN MANG SỐ NGUYÊN DƯƠNG

\(\Rightarrow x;y;z\in\varnothing\)

d) \(\left|x+\frac{1}{5}\right|+\left|3-y\right|=0\)

\(\Rightarrow\left|x+\frac{1}{5}\right|=\left|3-y\right|=0\)

\(\Rightarrow\left|x+\frac{1}{5}\right|=0\)                                \(\Rightarrow\left|3-y\right|=0\)

\(x+\frac{1}{5}=0\)                                                 \(3-y=0\)

\(x=\frac{-1}{5}\)                                                              \(y=3\)

VẬY X= -1/5; Y=3

CHÚC BN HỌC TỐT!!!!!!!

Ta có : 

\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=0-\frac{-3}{4}-\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=\frac{11}{20}\end{cases}}\)

Vậy \(x=\frac{-3}{4};y=\frac{1}{5};z=\frac{11}{20}\)

Giúp mình nhé các bạn mình đang cần gấp lắm

|5\(x\) - 4| = |\(x+2\)|

\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)

\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}

|2\(x\) - 3| - |3\(x\) + 2| = 0

|2\(x\) - 3| = | 3\(x\) + 2|

\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)

vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}

1. 

$(3^2-2^3)x+3^2.2^2=4^2.3$

$\Leftrightarrow x+36=48$

$\Leftrightarrow x=48-36=12$

2.

$x^5-x^3=0$

$\Leftrightarrow x^3(x^2-1)=0$

$\Leftrightarrow x^3(x-1)(x+1)=0$

$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.

$(x-1)^2+(-3)^2=5^2(-1)^{100}$

$\Leftrightarrow (x-1)^2+9=25$

$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$

$\Rightarrow x-1=4$ hoặc $x-1=-4$

$\Leftrightarrow x=5$ hoặc $x=-3$

4.

$(2x-1)^2-(2x-1)=0$

$\Leftrightarrow (2x-1)(2x-1-1)=0$

$\Leftrightarrow (2x-1)(2x-2)=0$

$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$

$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$

$\Lef

`@` `\text {Ans}`

`\downarrow`

\((3^2-2^3)x+3^2.2^2=4^2.3\)

`=> x + (3*2)^2 = 48`

`=> x+6^2 = 48`

`=> x + 36 = 48`

`=> x = 48 - 36`

`=> x=12`

Vậy, `x=12`

\(x^5-x^3=0\)

`=> x^3(x^2 - 1)=0`

`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy, `x \in {0; +- 1 }`

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)

`=> (x-1)^2 + 9 = 25*1`

`=> (x-1)^2 + 9 = 25`

`=> (x-1)^2 = 25 - 9`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

\((2x-1)^2-(2x-1)=0\)

`=> (2x-1)(2x-1) - (2x-1)=0`

`=> (2x-1)(2x-1-1)=0`

`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

bộ định không làm bài tập về nhà à , thấy bài cái là lên hỏi

có làm nhưng mà quên cách òi giúp cái coi

a,

\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)

d,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)

Bạn mới hỏi ở dưới rồi :v