Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

Vì \(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

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

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

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

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Bài 4 xem lại đề nhé bác

\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)

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

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

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

=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)

=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)

Vậy \(x\in\left\{\frac{5}{8}\right\}\)

\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)

=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)

=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)

=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)

=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)

Vậy \(x\in\left\{-\frac{39}{40}\right\}\)

\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)

=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)

=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)

( chiệt tiêu )

=> \(5x-6x+26=-14-7x\)

=> \(-x+26=-14-7x\)

=> \(-x+7x=-14-26\)

=> \(6x=-40\)

=> \(x=-40:6=\frac{20}{3}\)

Vậy \(x\in\left\{\frac{20}{3}\right\}\)

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

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

( chiệt tiêu )

=> \(2\left(2x-3\right)-9=5-3x-2\)

=> \(4x-6-9=3-3x\)

=> \(4x-15=3-3x\)

=> \(4x+3x=3+15\)

=> \(7x=18\)

=> \(x=18:7=\frac{18}{7}\)

Vậy \(x\in\left\{\frac{18}{7}\right\}\)

\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)

ĐKXĐ : \(x\ne0\)

=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)

=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)

=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)

=> \(\frac{32}{3x}=\frac{1}{4}\)

=> \(3x=32.4:1=128\)

=> \(x=128:3=\frac{128}{3}\)

Vậy \(x\in\left\{\frac{128}{3}\right\}\)

\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)

ĐKXĐ :\(x\ne1;\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)

=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)

=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)

=> \(\frac{26+5-2}{2\left(x-1\right)}\)

=> \(\frac{29}{2\left(x-1\right)}\)

\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)

=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)

=> \(x=\frac{19}{10}:2=\frac{19}{20}\)

Vậy \(x\in\left\{\frac{19}{20}\right\}\)

\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)

=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)

=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)

=> \(x=\frac{1}{2}:2=\frac{1}{4}\)

Vậy \(x\in\left\{\frac{1}{4}\right\}\)