ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\\x-3\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)

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

\(\frac{3\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}-\frac{2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}-\frac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

\(3x-9-2x+4-x+1=0\)

\(0x-4=0\Rightarrow0x=4\Rightarrow\) Phương trình vô nghiệm

a/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)

\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow3x=-15\Rightarrow x=-5\)

b/ ĐKXĐ: \(x\ne\left\{-\frac{4}{3};1\right\}\)

\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)

\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)

\(\Leftrightarrow44x=-33\Rightarrow x=-\frac{3}{4}\)

c/ ĐKXĐ: \(x\ne\left\{-\frac{1}{4};0\right\}\)

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

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

TH1: \(x^2-1=0\Rightarrow x=\pm1\)

TH2: \(\frac{3}{4x+1}-\frac{2}{x}-1=0\Leftrightarrow3x-2\left(4x+1\right)-x\left(4x+1\right)=0\)

\(\Leftrightarrow4x^2+6x+2=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)

a) \(3\left(x-4\right)+5=2\left(x+1\right)-8\)

\(\Leftrightarrow3x-12+5=2x+2-8\)

\(\Leftrightarrow x=1\)

Vậy : \(S=\left\{1\right\}\)

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

\(\Leftrightarrow5x^2+10x+5+2x=5x^2-3\)

\(\Leftrightarrow12x=-8\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy : \(S=\left\{-\frac{2}{3}\right\}\)

c) \(\frac{4\left(x+2\right)}{15}=\frac{13x-9}{40}\)

\(\Leftrightarrow32\left(x+2\right)=3\left(13x-9\right)\)

\(\Leftrightarrow32x-39x=-27-64\)

\(\Leftrightarrow-7x=-91\)

\(\Leftrightarrow x=13\)

Vậy : \(S=\left\{13\right\}\)

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

mà \(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

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

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)

a/ ĐKXĐ: \(x\ne-1\)

\(\Leftrightarrow4\left(3-7x\right)=x+1\)

\(\Leftrightarrow12-28x=x+1\)

\(\Rightarrow29x=11\Rightarrow x=\frac{11}{29}\)

b/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

\(\Leftrightarrow1-\left(\sqrt{x}-2\right)=3-\sqrt{x}\)

\(\Leftrightarrow3=3\) (luôn đúng)

Vậy nghiệm của pt là \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ne7\)

\(\Leftrightarrow8-x-8\left(x-7\right)=1\)

\(\Leftrightarrow8-x-8x+56=1\)

\(\Leftrightarrow-9x=-63\Rightarrow x=7\left(ktm\right)\)

Vậy pt vô nghiệm

d/ ĐKXĐ: \(x\ne4\)

\(\Leftrightarrow\frac{28}{6\left(x-4\right)}-\frac{6\left(x+2\right)}{6\left(x-4\right)}=\frac{-9}{6\left(x-4\right)}-\frac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

\(\Rightarrow x=5\)

e/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)

\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow3x=-15\Rightarrow x=-5\)

Nguyễn Việt Lâm giúp mk nhá, thenk kiu