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\(3\left(x-2\right)+4=5x-2\left(x-1\right)\\ \Leftrightarrow3x-6+4=5x-2x+2\\ \Leftrightarrow0x=4\left(vôlý\right)\)

Vậy pt vô nghiệm

\(2\left(x-2\right)-3\left(1-2x\right)=5\\ \Leftrightarrow2x-4-3+6x=5\\ \Leftrightarrow8x=12\\ \Leftrightarrow x=\dfrac{3}{2}\)

- Ta có: \(\left(x^2-1\right).\left(x+2\right).\left(x-3\right)=\left(x-1\right).\left(x^2-4\right).\left(x+5\right)\)

      \(\Leftrightarrow\left(x-1\right).\left(x+1\right).\left(x+2\right).\left(x-3\right)=\left(x-1\right).\left(x-2\right).\left(x+2\right).\left(x+5\right)\)

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

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

      \(\Leftrightarrow\left(x-1\right).\left(x+2\right).\left[\left(x^2-2x-3\right)-\left(x^2+3x-10\right)\right]=0\)

      \(\Leftrightarrow\left(x-1\right).\left(x+2\right).\left(x^2-2x-3-x^2-3x+10\right)=0\)

      \(\Leftrightarrow\left(x-1\right).\left(x+2\right).\left(-5x+7\right)=0\)

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

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

+ \(-5x+7=0\)\(\Leftrightarrow\)\(-5x=-7\)\(\Leftrightarrow\)\(x=\frac{7}{5}\left(TM\right)\)

Vậy \(S=\left\{-2,1,\frac{7}{5}\right\}\)

Đề sai thì phải, bạn thêm dấu ngoặc vào đi. Như vậy dễ làm hơn.

khong sai de'

Giải:

Ta có: \(\frac{1}{2}\left(x+1\right)\) + \(\frac{1}{4}\left(x+3\right)\) = \(3-\frac{1}{3\left(x-2\right)}\)

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

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

⇔

Còn lại bạn tự làm nhá!