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

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

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

\(\Leftrightarrow x=\dfrac{5}{2}\)

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

\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)

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

\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)

\(\Leftrightarrow x=\dfrac{1}{8}\)

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

\(=\left(2.\left(-\frac{1}{2}\right)+1\right)^2-3\left(-\frac{1}{2}-1\right)^2-\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)\)

\(=-3\left(-\frac{9}{4}\right)-\frac{1}{2}.\left(-\frac{3}{2}\right)\)

\(=\frac{27}{4}+\frac{3}{4}=\frac{31}{4}\)

còn đâu tự lm nha ! 

mn giúp e sớm đc ko ạ

\(\dfrac{2}{36a^2b^2-1}=\dfrac{2}{\left(6ab-1\right)\left(6ab+1\right)}\\ \dfrac{1}{6ab+1}=\dfrac{6ab-1}{\left(6ab-1\right)\left(6ab+1\right)};\dfrac{1}{6ab-1}=\dfrac{6ab+1}{\left(6ab-1\right)\left(6ab+1\right)}\)

\(\dfrac{x}{x^3-27}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{2x}{x^2-6x+9}=\dfrac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{1}{x^2+3x+9}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

\(\dfrac{x^2-x}{x^2-1}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2}\\ \dfrac{3x}{x^3+2x^2+x}=\dfrac{3x}{x\left(x^2+2x+1\right)}=\dfrac{3}{\left(x+1\right)^2}\\ 2x=\dfrac{2x\left(x+1\right)^2}{\left(x+1\right)^2}\)

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

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

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

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

\(\left(2x^2+5x+3\right):\left(x+1\right)-\left(4x-5\right)\)

\(=\dfrac{2x^2+2x+3x+3}{x+1}-4x+5\)

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

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

\(=2x+3-4x+5\)

\(=-2x+8\)

thay x=-2 vào biểu thức ta có:

\(=-2\left(-2\right)+8=4+8=12\)

Rối quá bn ơi

Bn ko dùng dấu enter để xuống dòng à 

Mk sẽ giúp bn 3 câu đầu thôi còn 3 câu sau thì..... chịu vì ko biết làm sao

a) \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)

(=)\(\left[9\left(2x+1\right)-4\left(x+1\right)\right]\left[9\left(2x+1\right)+4\left(x+1\right)\right]=0\)

(=)\(\left(18x+9-4x-4\right)\left(18x+9+4x+4\right)=0\)

(=)\(\left(14x+5\right)\left(22x+12\right)=0\)

(=)\(\left(14x+5\right)=0\) hoặc \(\left(22x+12\right)=0\)

1) \(14x+5=0\Rightarrow x=-\dfrac{5}{14}\)

2) \(22x+12=0\Rightarrow x=-\dfrac{6}{11}\)

Vậy........

b)\(\left(x+1\right)^2+2\left(x+1\right)+1=0\)

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

(=) \(\left(x+2\right)^2=0\)

(=) \(x+2=0\Rightarrow x=-2\)

Vậy........

c) \(\left(x-1\right)\left(x^2-9\right)+x+3=0\)

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

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

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

(=) \(x+3=0\) hoặc \(x-2=0\) hoặc \(x-1=0\)

1)\(x+3=0\Rightarrow x=-3\)

2)\(x-2=0\Rightarrow x=2\)

3) \(x-1=0\Rightarrow x=1\)

Vậy.........