6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

1+2+4+6+7+8+9+x = 8*5 +12x

=> 12x-x = 1+2+4+6+7+8+9 + 8*5

=> 11x = 77

=> x = 7

canhj hình vuông đó là 75

cạnh hình vuông đó là :75cm

cạnh = 75cm

ban co the viet lai dau / la phan so hay la dau chia

phân số

<=> 15/12x - 6/5x = -1/2 - 3/7

<=> 1/20x            = -13/14

<=>    x                = -13/14 : 1/20

<=>    x                = -130/7

Ta có 2(x+3)=3(1-x)-2

suy ra : 2x + 6 = 3 -3x -2

            2x +6 = 1-3x

            2x +6 -1 +3x=0

            5x+5 =0

           5(x+1)=0

    Vậy x bằng -1

bí rùi, nếu chỉ có 1 cái x thì mình giải được

Sai đề?

a) \(-12\left(x-5\right)+7\left(3-x\right)=5\)

\(-12x+60+21-7x=5\)

\(-19x=5-60-21\)

\(-19x=-76\)

\(x=4\)

vậy \(x=4\)

tương tự 

Yêu cầu?

1) \(x^2+8x+7=0\)

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

\(\Leftrightarrow x\left(x+7\right)+\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-1\end{matrix}\right.\)

Vậy...

2) \(4x^2+12x+6=0\)

\(\Leftrightarrow4\left(x^2+3x+\frac{3}{2}\right)=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{3}{4}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=\frac{3}{4}=\left(\frac{\pm\sqrt{3}}{2}\right)^2\)

\(\Leftrightarrow x=\frac{\pm\sqrt{3}-3}{2}\)

Vậy...

\(\left(4x+1\right)\left(12x-1\right)\left(3x-2\right)\left(x+1\right)-4\) (Sửa đề)

\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt \(12x^2+11x-1=n\)

\(=\left(n+3\right)n-4\)

\(=n^2+3n-4\)

\(=n^2-n+4n-4\)

\(=n\left(n-1\right)+4\left(n-1\right)\)

\(=\left(n-1\right)\left(n+4\right)\)

\(=\left(12x^2+11x-1-1\right)\left(12x^2+11x-1+4\right)\)

\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

\(\left(3x+4\right)\left(x+1\right)\left(6x+7\right)^2=6\)

\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)(1)

Đặt \(\left(3x^2+7x+4\right)=n\)lúc đó (1):

\(\left(12n+1\right)n=6\)

\(\Rightarrow\hept{\begin{cases}n=0,75\\n=\frac{2}{3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)