a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)

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

\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)

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

\(\Leftrightarrow x\left(-x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: S={0;-8}

b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)

c) Ta có: \(x^2+6x+9=4x^2\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: S={3;-1}

d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)

a) Quy đồng bỏ mẫu rồi giai pt ta đc : \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

b)\(x=1\)

Con bai 2 thi sao a

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!

b: =>15-x=-10

hay x=25

a: =>-2x+17=9

=>-2x=-8

hay x=4

d: \(\Leftrightarrow9x^2=81\)

hay \(x\in\left\{3;-3\right\}\)

e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)

Lằng nhằng vậy😅

các bn oi dau / nay la gia tri tuyet doi day

a) 13 - ( 2x - 11 ) = -15

=>       2x - 11    = 13 - ( -15 )

=>       2x - 11    = 28

=>      2x           = 28 + 11

=>     2x            = 39

=>      x            = 39 : 2

=>      x            = 19,5

Mấy câu khác bn tự giải nha!Cũng tương tự như vậy thôi!

a/ \(13-\left|2x-11\right|=-15\)

\(\Leftrightarrow\left|2x-11\right|=28\)

\(\Leftrightarrow\orbr{\begin{cases}2x-11=28\\2x-11=-28\end{cases}}\) 

\(\Leftrightarrow\orbr{\begin{cases}x=19,5\\x=-8,5\end{cases}}\)

Vậy ...

b/ \(17-\left(13-2x\right)=x+11\)

\(\Leftrightarrow17-13+2x=x+11\)

\(\Leftrightarrow5+2x=x+11\)

\(\Leftrightarrow2x-x=11-5\)

\(\Leftrightarrow x=6\)

Vậy ,,,

b/ \(17-\left|3x-2\right|=7\)

\(\Leftrightarrow\left|3x-2\right|=10\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=10\\3x-2=-10\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=12\\3x=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{8}{3}\end{cases}}\)

Vậy ...

d/ \(\left|4x\right|=2x+2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+2=4x\\2x+2=-4x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-4x=2\\2x+4x=2\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

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

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

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

$\left({}^{}-2\right)\left({}^{}-2x+2\right)\left({}^{}+2x+2\right)\left({}^{}+2\right)$

$=\left[\left({}^{}-2\right)\left({}^{}+2\right)\right].\left\{\left[\left({}^{}+2\right)-2x\right].\left[\left({}^{}+2\right)+2x\right]\right\}$

$=\left({}^{}-4\right).\left[{}^{}-4{}^{}\right]$

$=\left({}^{}-4\right)\left({}^{}+4{}^{}+4-4{}^{}\right)=\left({}^{}-4\right)\left({}^{}+4\right)$

$={}^{}-1$

Trả lời:

( x² - 2x + 2 ) ( x² - 2 ) ( x² + 2x + 2 ) ( x² + 2 )

= [ ( x² - 2x + 2 ) ( x² + 2x + 2 ) ] [ ( x² - 2 ) ( x² + 2 ) ]

= [ ( x² + 2 )² - ( 2x )² ] ( x⁴ - 4 )

= ( x⁴ + 4x² + 4 - 4x² ) ( x⁴ - 4 )

= ( x⁴ + 4 ) ( x⁴ - 4 )

= ( x⁴ )² - 4²

= x⁸ - 16

Bài4:

\(a,A=2x\left(x-1\right)-x\left(2x+1\right)-\left(3-3x\right)\\ =2x^2-2x-2x^2-x-3+3x\\ =-3\)

Vậy...(đpcm)

\(b,B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\\ =2x^2-6x-2x^2+6x-4\\ =-4\)

Vậy...(đpcm)

\(c,C=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\\ =6x^2+23x-55-6x^2-23x-21\\ =-21\)

Vậy...(đpcm)

d,Câu d là câu c

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

\(A=2x^2-2x-2x^2-x-3+3x\)

\(A=-3\)

\(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)

\(B=2x^2-6x-2x^2+4x-2x+4x-4\)

\(B=-4\)