bài 1

a) 155 - 10.(x+1) = 55

=>10 .(x+1) = 100

=>x + 1 = 10

=>x = 9

còn lại tương tự 

Tính nhanh:

17/5*-31/125*1/2*10/17*-1/2^3

a)125 : x = 2² - (-1)

125 : x = 4 + 1

125 : x = 5

x = 125 : 5

x = 25

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b) 2x - 8 = -4

2x = (-4) + 8

2x = 4

x = 4 : 2 

x = 2

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c) Xem lại đề.

\(125:x=2^2-\left(-1\right)\)

\(=>125:x=4+1\)

\(=>125:x=5\)

\(=>x=125:5\)

\(=>x=25\)

_____

\(2x-8=-4\)

\(=>2x=\left(-4\right)+8\)

\(=>2x=4\)

\(=>x=4:2\)

\(=>x=2\)

_______

\(6^{2x+5}=216\)

\(=>6^{2x+5}=6^3\)

\(=>2x+5=3\)

\(=>2x=3-5\)

\(=>2x=-2\)

\(=>x=\left(-2\right):2\)

\(=>x=-1\)

\(#NqHahh\)

1: \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>23x+61=2x+61

hay x=0

2: \(\dfrac{6}{x-5}+\dfrac{x+2}{x-8}=\dfrac{18}{\left(x-5\right)\left(8-x\right)}-1\)

\(\Leftrightarrow6x-48+x^2-3x-10=-18-x^2+13x-40\)

\(\Leftrightarrow x^2+3x-58+x^2-13x+58=0\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0

c: \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)

\(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)=-7x^2+3x\)

\(\Leftrightarrow x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x=0\)

\(\Leftrightarrow x^2=0\)

hay x=0

bài của p hay trog sgk
 

5: =>4x^2-1/9=0

=>(2x-1/3)(2x+1/3)=0

=>x=1/6 hoặc x=-1/6

6: =>x-1=2

=>x=3

7:=>(2x-1)^3=-27

=>2x-1=-3

=>2x=-2

=>x=-1

8: =>1/8(x-1)^3=-125

=>(x-1)^3=-1000

=>x-1=-10

=>x=-9

3: =>(5x-5)^2-4=0

=>(5x-7)(5x-3)=0

=>x=3/5 hoặc x=7/5

4: =>(5x-1)^2=0

=>5x-1=0

=>x=1/5

1: =>(3x-1)(2x-1)=0

=>x=1/3 hoặc x=1/2

2: =>x^2(2x-3)-4(2x-3)=0

=>(2x-3)(x^2-4)=0

=>(2x-3)(x-2)(x+2)=0

=>x=3/2;x=2;x=-2

`@` `\text {Answer}`

`\downarrow`

`1,`

\(2x\left(3x-1\right)+1-3x=0\)

`<=> 2x(3x - 1) - 3x + 1 = 0`

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

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

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

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

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

Vậy,  `S = {1/2; 1/3}`

`2,`

\(x^2\left(2x-3\right)+12-8x=0\)

`<=> x^2(2x - 3) - 8x + 12 =0`

`<=> x^2(2x - 3) - (8x - 12) = 0`

`<=> x^2(2x - 3) - 4(2x - 3) = 0`

`<=> (x^2 - 4)(2x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy, `S = {+-2; 3/2}`

`3,`

\(25\left(x-1\right)^2-4=0\)

`<=> 25(x-1)(x-1) - 4 = 0`

`<=> 25(x^2 - 2x + 1) - 4 = 0`

`<=> 25x^2 - 50x + 25 - 4 = 0`

`<=> 25x^2 - 15x - 35x + 21 = 0`

`<=> (25x^2 - 15x) - (35x - 21) = 0`

`<=> 5x(5x - 3) - 7(5x - 3) = 0`

`<=> (5x - 7)(5x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy, `S = {7/5; 3/5}`

`4,`

\(25x^2-10x+1=0\)

`<=> 25x^2 - 5x - 5x + 1 = 0`

`<=> (25x^2 - 5x) - (5x - 1) = 0`

`<=> 5x(5x - 1) - (5x - 1) = 0`

`<=> (5x - 1)(5x-1)=0`

`<=> (5x-1)^2 = 0`

`<=> 5x - 1 = 0`

`<=> 5x = 1`

`<=> x = 1/5`

Vậy,` S = {1/5}.`