Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m

Bài 2:

a) \(x+x^2=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)

b) \(0x-3=0\)

\(\Leftrightarrow0x=3\)

\(\Rightarrow vonghiem\)

c) \(3y=0\)

\(\Leftrightarrow y=0\)

Bài 1 dài nên lười làm quá :)

\(\frac{1}{3}\left(3+\frac{3}{5}x\right)-4x=20\%x-1\)

=> \(1+\frac{1}{5}x-4x=\frac{1}{5}x-1\)

=> \(1+\frac{1}{5}x-4x-\frac{1}{5}x+1=0\)

=> \(\left(1+1\right)+\left(\frac{1}{5}x-\frac{1}{5}x-4x\right)=0\)

=> \(2-4x=0\)

=> \(4x=2\)

=> \(x=\frac{1}{2}\)

Vậy : ...

P/S : Lớp 6 có phương trình ???

\(\frac{1}{3}\left(3+\frac{3}{5}x\right)-4x=20\%.x-1\)

\(\Leftrightarrow1+\frac{1}{5}x-4x=\frac{1}{5}x-1\)

\(\Leftrightarrow1-4x=-1\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=2\)

a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)

\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)

\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)

\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)

tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi 

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

\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)

\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)

\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)

\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)

\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)

\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)

\(\Leftrightarrow23x=69\)

\(\Leftrightarrow x=3\)

Vậy nghiệm của pt x=3

a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)

\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)

\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)

\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)

\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)

\(< =>3072-107x=\frac{38x-684}{5}\)

\(< =>\left(3072-107x\right)5=38x-684\)

\(< =>15360-535x-38x-684=0\)

\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)

nghệm xấu thế 

\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)

\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)

\(< =>993-33x-11x-415=0\)

\(< =>578=44x< =>x=\frac{289}{22}\)

Bài 1: 

b) Phương trình đã cho tương đương với phương trình:

\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)

\(\Leftrightarrow44x=-1056\)

\(\Leftrightarrow x=-24\)

Vậy x=-24 là nghiệm của phương trình

c) Phương trình đã cho tương đương với phương trình:

\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)

\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)

\(\Leftrightarrow-447x=894\)

\(\Leftrightarrow x=-2\)

Vậy x=-2 là nghiệm của phương trình