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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 \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
a) \(\frac{5-2x}{3}+\frac{\left(x+1\right)\left(x-1\right)}{3x-1}=\frac{\left(x+2\right)\left(1-3x\right)}{9x-3}\)
<=> \(\frac{5-2x}{3}+\frac{\left(x+1\right)\left(x-1\right)}{3x-1}=-\frac{\left(x+3\right)\left(3x-1\right)}{3\left(3x-1\right)}\)
<=> \(\frac{5-2x}{3}+\frac{\left(x+1\right)\left(x-1\right)}{3x-1}=-\frac{x+2}{3}\)
<=> (5 - 2x)(3x - 1) + 3(x + 1)(x - 1) = -(x + 2)(3x - 1)
<=> 15x - 5 - 6x^2 + 2x + 3x^2 - 3x + 3x - 3 = -3x^2 - 6x + x + 2
<=> 17x - 8 = -5x + 2
<=> 17x - 8 + 5x = 2
<=> 22x - 8 = 2
<=> 22x = 2 + 8
<=> 22x = 10
<=> x = 10/22 = 5/11
b) \(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)
<=> 2(x - 1) + (x - 5)(x - 3) = (x - 3)(x - 1)
<=> 2x - 2 + x^2 - 3x - 5x + 15 = x^2 - x - 3x + 3
<=> -6x + 13 = -4x + 3
<=> -6x + 13 + 4x = 3
<=> -2x + 13 = 3
<=> -2x = 3 - 13
<=> -2x = -10
<=> x = 5
a) \(\frac{5-2x}{3}+\frac{\left(x-1\right)\left(x+1\right)}{3x-1}=\frac{\left(x+2\right)\left(1-3x\right)}{9x-3}\left(x\ne\frac{1}{3}\right)\)
\(\Leftrightarrow\frac{5-2x}{3}+\frac{\left(x-1\right)\left(x+1\right)}{3x-1}-\frac{\left(x+2\right)\left(1-3x\right)}{3\left(3x-1\right)}=0\)
<=> \(\frac{\left(5-2x\right)\left(3x-1\right)}{3\left(3x-1\right)}+\frac{3\left(x^2-1\right)}{3\left(3x-1\right)}-\frac{x-3x^2+2-6x}{3\left(3x-1\right)}=0\)
<=> \(\frac{15x-5-6x^2+2x}{3\left(3x-1\right)}+\frac{3x^2-3}{3\left(3x-1\right)}-\frac{-3x^2-5x+2}{3\left(3x-1\right)}=0\)
<=> \(\frac{-6x^2+17x-5}{3\left(3x-1\right)}+\frac{3x^2-3}{3\left(3x-1\right)}-\frac{-3x^2-5x+2}{3\left(3x-1\right)}=0\)
<=> \(\frac{-6x^2+17x-5+3x^2-3+3x^2+5x-2}{3\left(3x-1\right)}=0\)
<=> \(\frac{22x-10}{3\left(3x-1\right)}=0\)
=> 22x-10=0
<=> \(x=\frac{5}{11}\)(tmđk)