giup mik voi

$3x-10=14$

$3x=14+10$

$3x=24$

$x=24:3$

$x=8$

Lời giải:

$(7x-12)+(8x-10)+(3x-14)=4$

$(7x+8x+3x)-(12+10+14)=4$
$18x-36=4$

$18x=40$

$x=\frac{20}{9}$

\(7x-12+8x-10+3x-14-4=0\)

\(18x-40=0\)

\(x=\dfrac{40}{18}=\dfrac{20}{9}\)

a) (2y - 5)² - 2y(2y - 10)

= 4y² - 20y + 25 - 4y² + 20y

= 25

b) (3x - 4)(3x + 4) - [(-3x)² -14]

= 9x² - 16 - 9x² + 14

= - 2

câu a) bạn giải sai thì phải

a: \(\dfrac{x^2-5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\)

\(=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)

\(=\dfrac{x\left(x-3\right)}{\left(x-2\right)\left(x+4\right)}\)

b: \(\dfrac{x^2+2x-3}{x^2+3x-10}:\dfrac{x^2+7x+12}{x^2-9x+14}\)

\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x+4\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)

***a, 14+2x=-32

        2x=-32-14

        2x=-46

          x=-46/2

         x=-23

        Vậy x=-23

ko biết ahihi

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xy+x+y=2

xy+x+y+1=2+1

(xy+x)+(y+1)=3

x(y+1)+(y+1)=3

(x+1)(y+1)=3=1.3=3.1=-1.-3=-3.-1

\(\Rightarrow\left[{}\begin{matrix}x+1=1;y+1=3\\x+1=3;y+1=1\\x+1=-1;y+1=-3\\x+1=-3;y+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0;y=2\\x=2;y=0\\x=-2;y=-4\\x=-4;y=-2\end{matrix}\right.\)

Vậy:.................

xy+14+2y+7x= -10

\(\Leftrightarrow\)y(x+2)+7(x+2)=-10

\(\Leftrightarrow\)(y+7)(x+2)=-10=1.(-10)=2.(-5)=5.(-2)=10.(-1)

3x-(-17)=14

3x=14+(-17)

3x=-3

x=(-3):3

x=-1

Vậy x=-1.

|x+9|.2=10

|x+9|=10:2

|x+9|=5

\(\Rightarrow\orbr{\begin{cases}x+9=-5\\x+9=5\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\x=-4\end{cases}}}\)

Vậy \(x\in\left\{-14;-4\right\}\)

\(3x-\left(-17\right)=14\)

\(3x+17=14\)

\(3x=14-17\)

\(3x=-3\)

\(x=-1\)

\(a,\left|x+9\right|.2=10\)

\(\left|x+9\right|=5\)

\(th1:x+9=5\)

\(x=-4\)

\(Th2:x+9=-5\)

\(x=-14\)

Vậy.....

hok tốt!!