`a)|2x-15|=13`

`**2x-15=13`

`<=>2x=28`

`<=>x=14.`

`**2x-15=-13`

`<=>2x=-2`

`<=>x=-1.`

`b)|7x+3|=66`

`**7x+3=66`

`<=>7x=63`

`<=>x9`

`**7x+3=-66`

`<=>7x=-69`

`<=>x=-69/7`

`c)|5x-2|=0`

`<=>5x-2=0`

`<=>5x=2`

`<=>x=2/5`

\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

Vậy ...

\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

Vậy ...

\(c,\Leftrightarrow5x-2=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy ...

a \(\Rightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=18\\2x=-8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

b \(\Rightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}7x=63\\7x=-69\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

c \(\Rightarrow5x-2=0\Rightarrow x=\dfrac{2}{5}\)

a) I2x-5I=13

<=> 2x-5=13 hoặc 2x-5=-13

<=> 2x=18 hoặc 2x=-8

<=> x=9 hoặc x=-4

b) I7x+3I=66

<=> 7x+3=66 hoặc 7x+3=-66

<=> 7x=63 hoặc 7x=-69

<=> x=9 hoặc x=\(\frac{-69}{7}\)

a) \(\left|2x-5\right|=13\)

\(\Leftrightarrow2x-5=\pm13\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=13\\2x-5=-13\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=18\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\2x=-4\end{cases}}}\)

Vậy \(x\in\left\{9;-4\right\}\)

b) \(\left|7x+3\right|=66\)

\(\Leftrightarrow7x+3=\pm66\)

\(\Leftrightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=63\\7x=-69\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=\frac{-69}{7}\end{cases}}}\)

Vậy \(x\in\left\{9;\frac{-69}{7}\right\}\)

a . | 2x - 5 | = 13

\(\Rightarrow\orbr{\begin{cases}2x-5=13\\2x-5=-13\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=13+5\Rightarrow x=18:2=9\\2x=-13+5\Rightarrow x=-8:2=-4\end{cases}}\)

Vậy .....................................................................

b. | 7x + 3 | = 66

 \(\Rightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}7x=63\Rightarrow x=63:7=9\\7x=-69\Rightarrow x=-69:7=-\frac{69}{7}\end{cases}}\)

vậy....................................................................................

a) 2x(x - 5) - 2x² = 2x² - 10x - 2x² = -10x = 20 => x = 20 : (-10) = -2

b) 5x(2x - 7) + 2x(8 - 5x) = 10x² - 35x + 16x - 10x² = -19x = 5 => x = \(\frac{5}{-19}=\frac{-5}{19}\)

c) 4x(7x - 5) - 7x(4x - 2) = 28x² - 20x - 28x² + 14x = -6x = -12 => x = -12 : (-6) = 2

MK mới học lớp 7 thôi nhưng mk làm vài câu nha

a) 2x(x-5)-2x²=20

    2x²-100x-2x²=20

    100x=20

     x=20:100

      x=\(\frac{1}{5}\)

a) Ta có : 2x ( x - 5 ) - 2x² = 20

=> 2x ( x - 5 ) - 2x.x = 20

=> 2 x ( x - 5 - x ) = 20

=> 2x . ( -5 ) = 20

=> -10 x = 20

=> x = -2

Chọn mik mik làm tiếp

         \(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

\(x^3-5x^2+x-5=0\)

\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

7 \(\times\) ( 2\(x\) - 5) - 5 \(\times\) (7\(x\) - 2) + 2 \(\times\) (5\(x\) - 7) = (\(x\) - 2) - (\(x\) +4)

14\(x\) - 35 - 35\(x\) + 10 + 10\(x\) - 14 = \(x\) - 2 - \(x\) - 4

(14\(x\) - 35\(x\) + 10\(x\)) - (35 - 10+ 14)  = -6

(- 21 \(x\) + 10\(x\)) - (25 + 14) = - 6

         -11\(x\) - 39 = - 6

          -11\(x\)       = - 6 + 39

          - 11\(x\)     = 33

                \(x\) = 33 : (-11)

                \(x\) =  - 3

ko đúng r bạn ơi

bạn xem đề bài đúng chưa? Vế phải là phép nhân hay phép trừ?

Câu 1: 

a) Ta có: x-3 là ước của 13

\(\Leftrightarrow x-3\inƯ\left(13\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{4;2;16;-10\right\}\)(thỏa mãn)

Vậy: \(x\in\left\{4;2;16;-10\right\}\)

b) Ta có: \(x^2-7\) là ước của \(x^2+2\)

\(\Leftrightarrow x^2+2⋮x^2-7\)

\(\Leftrightarrow x^2-7+9⋮x^2-7\)

mà \(x^2-7⋮x^2-7\)

nên \(9⋮x^2-7\)

\(\Leftrightarrow x^2-7\inƯ\left(9\right)\)

\(\Leftrightarrow x^2-7\in\left\{1;-1;3;-3;9;-9\right\}\)

mà \(x^2-7\ge-7\forall x\)

nên \(x^2-7\in\left\{1;-1;3;-3;9\right\}\)

\(\Leftrightarrow x^2\in\left\{8;6;10;4;16\right\}\)

\(\Leftrightarrow x\in\left\{2\sqrt{2};-2\sqrt{2};-\sqrt{6};\sqrt{6};\sqrt{10};-\sqrt{10};2;-2;4;-4\right\}\)

mà \(x\in Z\)

nên \(x\in\left\{2;-2;4;-4\right\}\)

Vậy: \(x\in\left\{2;-2;4;-4\right\}\)

Câu 2: 

a) Ta có: \(2\left(x-3\right)-3\left(x-5\right)=4\left(3-x\right)-18\)

\(\Leftrightarrow2x-6-3x+15=12-4x-18\)

\(\Leftrightarrow-x+9+4x+6=0\)

\(\Leftrightarrow3x+15=0\)

\(\Leftrightarrow3x=-15\)

hay x=-5

Vậy: x=-5

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

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