(x+2)^2-(x-2)(x+2)=0

=> (x+2)(x+2-x+2)=0

=> (x+2).4=0

=> x+2=0

=> x=-2

mấy câu còn lại tự làm nha

a) (x+2)^2-(x-2)(x+2)=0 

 (x+2).[x+2-x+2]=0

(x+2).4=0

 x+2=0

x=-2

 b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

   4x²-4x+1-4x²+25=18

   26-4x=18

   4x=8

    x=2

 c)( 2x - 1)^2 - 25 = 0

    ( 2x - 1)^2 - 5² = 0

     (2x-1-5)(2x-1+5)=0

    (2x-6)(2x+4)=0

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Làm rõ từng ý nhé các bạn !

Cảm ơn các bạn rất nhiều!

nhân đa thức trước bạn nhé!

<=>6x^2 -(6x^2 +4x -9x -6)-1=0

phía trước là dấu trừ nên đổi dấu hạng tử bên trong

<=> 6x^2 -6x^2 -4x +9x+6 -1=0

<=>5x =-5

<=> x=-1

\(6x^2-\left(2x-3\right).\left(3x+2\right)-1=0\) \(0\)

\(< =>6x^2+\left(-2x+3\right).\left(3x+2\right)-1=0\)

\(< =>6x^2-6x^2-4x+9x+6-1=0\)

\(< =>5x=-5\)

\(< =>x=-1\)

\(\frac{x+7}{3}+\frac{x+5}{4}=\frac{x+3}{5}+\frac{x+1}{6}\)

\(\Rightarrow\frac{x+7}{3}+2+\frac{x+5}{4}+2=\frac{x+3}{5}+2+\frac{x+1}{6}+2\)

\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}=\frac{x+13}{5}+\frac{x+13}{6}\)

\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}-\frac{x+13}{5}-\frac{x+13}{6}=0\)

\(\Rightarrow\left(x+13\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Vì \(\left(\frac{1}{3}>\frac{1}{4}>\frac{1}{5}>\frac{1}{6}\right)\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>0\)

\(\Rightarrow x+13=0\Leftrightarrow x=-13\)

\(\frac{x+m}{n+p}+\frac{x+n}{p+m}+\frac{x+p}{n+m}+3=0\)

\(\Rightarrow\frac{x+m}{n+p}+1+\frac{x+n}{p+m}+1+\frac{x+p}{n+m}+1=0\)

\(\Rightarrow\frac{x+m+n+p}{n+p}+\frac{x+m+n+p}{p+m}+\frac{x+m+n+p}{n+m}=0\)

\(\Rightarrow\left(x+m+n+p\right)\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)=0\)

Vì m,n,p là số dương nên \(\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)>0\)

\(\Rightarrow x+m+n+p=0\Rightarrow x=-\left(m+n+p\right)\)

\(\frac{5x+\frac{3x-4}{5}}{15}=\frac{\frac{3-x}{15}+7x}{5}+1-x\)

\(\Rightarrow\frac{\frac{25x+3x-4}{5}}{15}=\frac{\frac{3-x+105x}{15}}{5}+1-x\)

\(\Rightarrow\frac{\frac{28x-4}{5}}{15}=\frac{\frac{3+104x}{15}}{5}+1-x\)

\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x}{75}+1-x\)

\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x+75-75x}{75}\)

\(\Rightarrow\frac{28x-4}{75}=\frac{78+29x}{75}\)

\(\Rightarrow28x-4=78+29x\)

\(\Rightarrow x=-82\)

a) <=> |-5X| =3X +16 

DK : X >-16/3

-5X = 3X +16 HOAC -5X =-3X-16 

-8X = 16 HOAC -2X = -16 

X= -2 HOAC X= 8 

VẬY S= {-2; 8}

b) <=> 3X +X = 1+2

<=> 4X = 3

<=> X=3/4

VẬY S={3/4}   

c) DK : X> 10/4

-2X = 4X-10 HOAC -2X = -4X +10 

-6X = 10 HOAC 2X = 10

X= -5/3 (LOAI) HOAC X= 5 (NHAN)

VẬY S={5}

LƯU Ý: CÓ CHỮ " HOẶC" THÌ KHÔNG CẦN MŨI TÊN HAI CHIỀU 

-MÌNH CHỈ GHI CÁCH GIẢI THÔI NHÉ 

CHÚC BẠN HỌC TỐT .

giúp mình vớiiii

c)  \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)

\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)

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

e)  \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)