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a: \(x+7⋮x+2\)

=>\(x+2+5⋮x+2\)

=>\(5⋮x+2\)

=>\(x+2\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-1;-3;3;-7\right\}\)

b: \(2x+5⋮x+1\)

=>\(2x+2+3⋮x+1\)

=>\(3⋮x+1\)

=>\(x+1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{0;-2;2;-4\right\}\)

c: \(3x-2⋮x+3\)

=>\(3x+9-11⋮x+3\)

=>\(-11⋮x+3\)

=>\(x+3\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{-2;-4;8;-14\right\}\)

d: \(12x+1⋮3x+2\)

=>\(12x+8-7⋮3x+2\)

=>\(-7⋮3x+2\)

=>\(3x+2\in\left\{1;-1;7;-7\right\}\)

=>\(3x\in\left\{-1;-3;5;-9\right\}\)

=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{5}{3};-3\right\}\)

e: \(x^2+3x+5⋮x+3\)

=>\(x\left(x+3\right)+5⋮x+3\)

=>\(5⋮x+3\)

=>\(x+3\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-2;-4;2;-8\right\}\)

f: \(x^2-2x+3⋮x+2\)

=>\(x^2+2x-4x-8+11⋮x+2\)

=>\(11⋮x+2\)

=>\(x+2\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{-1;-3;9;-13\right\}\)

Câu a) -3 phần 1/2 

Câu a) 2 mũ 2

a: =>-7/2:(4/5-1/2x)=4

=>4/5-1/2x=-7/2:4=-7/8

=>1/2x=4/5+7/8=67/40

=>x=67/20

b: =>5x=5

=>x=1

c: =>x(-2/3-1/3)=-2

=>-x=-2

=>x=2

d: =>-2/3(x+1)=-1/3+1/2=1/6

=>x+1=-1/6:2/3=-1/6*3/2=-3/12=-1/4

=>x=-1/4-1=-5/4

a) Thay \(x=2\) vào phương trình, ta được:

 \(15\left(m+6\right)+12=80\) \(\Rightarrow m=-\dfrac{22}{15}\)

Vậy \(m=-\dfrac{22}{15}\)

b) Thay \(x=1\) vào phương trình, ta được:

  \(15\left(2+m\right)-32=43\) \(\Rightarrow m=3\)

Vậy \(m=3\)

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

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)

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

c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)

\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)

\(\Rightarrow x\left(6x-2-15-6x\right)\)

\(\Rightarrow-16x=0\)

\(\Rightarrow x=0\)

d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)

\(\Rightarrow9x^2-4-4x+4=0\)

\(\Rightarrow9x^2-4x=0\)

\(\Rightarrow x\left(9x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

\(c,2x\left(3x-1\right)-3x\left(5+2x\right)=0\\ \Leftrightarrow6x^2-2x-15x-6x^2=0\\ \Leftrightarrow-17x=0\\ \Leftrightarrow x=0\\ d,\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\\ \Leftrightarrow9x^2-4-4x+4=0\\ \Leftrightarrow9x^2-4x=0\\ \Leftrightarrow x\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)

\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)

\(\Leftrightarrow3x=3\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a) \(\left|2x-5\right|=x+1\)

<=> \(\orbr{\begin{cases}2x-5=x+1\left(x\ge\frac{5}{2}\right)\\5-2x=x+1\left(x< \frac{5}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-6\left(ktm\right)\\3x=4\end{cases}}\)

<=> \(x=\frac{4}{3}\left(tm\right)\)

b) \(\left|3x-2\right|-1=2x\) <=> \(\left|3x-2\right|=2x+1\)

<=> \(\orbr{\begin{cases}3x-2=2x+1\left(x\ge\frac{2}{3}\right)\\2-3x=2x+1\left(x< \frac{2}{3}\right)\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-3\left(ktm\right)\\5x=1\end{cases}}\) <=> \(x=\frac{1}{5}\left(tm\right)\)

c) \(\left|x-5\right|+5=x\) <=> \(\left|x-5\right|=x-5\)

<=> \(\orbr{\begin{cases}x-5=x-5\left(x\ge5\right)\\5-x=x-5\left(x< 5\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\\2x=10\end{cases}}\) <=> 0x = 0 (luôn đúng) hoặc x = 5 (ktm)

Vậy x \(\ge\)5

d) \(\left|3x-5\right|=3x-5\) <=> \(\orbr{\begin{cases}3x-5=3x-5\left(x\ge\frac{5}{3}\right)\\5-3x=3x-5\left(x< \frac{5}{3}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\left(luônđúng\right)\\6x=10\end{cases}}\)

<=> \(\orbr{\begin{cases}x\ge\frac{5}{3}\\x=\frac{5}{3}\left(ktm\right)\end{cases}}\)Vậy x \(\ge\)5/3

$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`