a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)

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

=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)

b) \(2x-\left|x+1\right|=\frac{1}{2}\)

=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))

=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)

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

=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)

mk cgx mắc bài này

a) \(\left(2x-1\right)^{10}=\left(1-2x\right)^5\)

\(\Rightarrow\left(2x-1\right)^2=1-2x\)

\(\Rightarrow4x^2-4x+1=1-2x\)

\(\Rightarrow4x^2-4x=-2x\)

\(\Rightarrow2x^2-2x=-x\)

\(\Rightarrow2x^2-x=0\)

\(\Rightarrow x.\left(2x-1\right)=0\)

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

Vậy ...

b) \(\left(3x-1\right)^{15}=\left(1-3x\right)^8\)

\(\Rightarrow\left(3x-1\right)^{15}-\left(1-3x\right)^8=0\)

\(\Rightarrow\left(3x-1\right)^{15}-\left(-\left(3x-1\right)\right)^8=0\)

\(\Rightarrow\left(3x-1\right)^{15}-\left(3x-1\right)^8=0\)

\(\Rightarrow\left(3x-1\right)^8.\left(\left(3x-1\right)^7-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^8=0\\\left(3x-1\right)^7-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ....

c) Tự lm

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)

\(\Leftrightarrow2x+3=93\)

\(\Leftrightarrow2x=90\)

\(\Leftrightarrow x=45\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(\Rightarrow2x=90\)

\(\Rightarrow x=45\)

Vậy x = 45.

a) Đặt 2 x - 15 = t

TA có :

    \(t^5=t^3\)  => \(t^5-t^3=0\Leftrightarrow t^3\left(t^2-1\right)=0\)

=> t^3 = 0 hoặc t^2 - 1 = 0 

=> t =0 hoặc t^2 = 1 

=> t = 0 hoặc t = 1 hoặc t = -1

(+) t = 0 => 2x - 15 = 0 => x = 15/2

(+) 2x-  15 = 1 => 2x = 16 => x = 8

(+) 2x-  1 5 = -1 => 2x = 14 => x = 7 

b) x^2 < 5 

=> x < \(\sqrt{5}\approx2,2\)

Vì x thuộc N => x  = { 0;1;2) 

a) (2x-15)⁵ = (2x - 15)³

=> 2x - 15 = 1; 2x - 15 = - 1 ; 2x - 15 = 0

TH1: 2x - 15 = 1

=> 2x = 15 + 1= 16 (chọn vì là STN)

x = 16 : 2 = 8

TH2: 2x - 15 = - 1

2x = -1 + 15 = 14

=> x = 14 : 2 = 7 (chọn vì là STN)

TH2: 2x - 15 = 0

2x = 0 + 15 = 15

=> x = 15: 2 = 7,5 (loai vì là số thập phân)

=> x = 7 ; hoặc x = 8            

a, \(\left|x+1\right|-4=0\)

\(\Rightarrow\left|x+1\right|=4\)

+) Xét \(x\ge-1\) có:

\(x+1=4\Rightarrow x=3\) ( t/m )

+) Xét x < -1 có:

\(-x-1=4\Rightarrow x=-5\) ( t/m )

Vậy x = 3 hoặc x = -5

b, tương tự

c, \(3x+\left|2x\right|=5x\)

\(\Rightarrow\left|2x\right|=2x\)

+) Xét \(x\ge0\)

\(\Rightarrow2x=2x\Rightarrow x\in R\forall x>0\)

+) Xét x < 0

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

Vậy \(x\ge0\)

d, \(\left|2x+1\right|=\left|x-3\right|\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=x-3\\2x+1=3-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy...

a, \(2x.\left(x-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)

b, \(\frac{2}{3}.\left(\frac{2}{5}+x\right)=\frac{2}{15}\)

\(\Leftrightarrow\frac{2}{5}+x=\frac{1}{5}\)

\(\Leftrightarrow x=-\frac{1}{5}\)

NX: 2x+3; 5(2x+3) và 2(2x+3)  cùng dấu

+TH1: 2x+3 \(\ge\)0 => x \(\ge\frac{-3}{2}\)

=> 5(2x+3), 2(2x+3) \(\ge\)0

=> |5(2x+3)| = 5(2x+3); |2(2x+3)| = 2(2x+3); |2x+3| = 2x+3

=> (2x+3)(5+2+1) = 16

=> 2x+3 = 2

=> 2x = -1

=> x = -1/2 (t/m)

+ TH2: 2x+3 < 0 => x < -3/2

cmtt => -5(2x+3) - 2(2x+3) - (2x+3) = 16

=> (2x+3)(-5-2-1) = 16

=> 2x+3 = -2

=> 2x = -5

=> x = -5/2 (t/m)

/8(2x+3/ = 16

/2x+3/=2

2x+3=2 hoặc 2x+3=-2

2x=-1 hoặc 2x=-5

x=-1/2 hoặc x=-5/2

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