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Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
Bài 2.
\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)
ĐK: \(x\ne2\)
\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)
ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
a) ĐKXĐ: \(x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> \(5+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> 5(x - 4)(x + 4) + 96(x - 4) = (2x - 1)(x - 4)(4 - x) - (3x - 1)(x + 4)(4 - x)
<=> 20x2 - 16x + 64 = 18x2 + 8x
<=> 20x2 - 16x + 64 - 18x2 - 8x = 0
<=> 2x2 - 24x + 64 = 0
<=> 2(x2 - 12x + 32) = 0
<=> 2(x - 8)(x - 4) = 0
<=> (x - 8)(x - 4) = 0
<=> x - 8 = 0 hoặc x - 4 = 0
<=> x = 8 (tm) hoặc x - 4 = 0 (ktm)
=> x = 8
b) ĐKXĐ: \(x\ne\pm\frac{2}{3}\)
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-2^2}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
<=> (2 + 3x)2 - 6(3x - 2) = 9x2
<=> 16 - 6x + 9x2 = 9x2
<=> 16 - 6x + 9x2 - 9x2 = 0
<=> 16 - 6x = 0
<=> -6x = 0 - 16
<=> -6x = -16
<=> x = -16/-6 = 8/3
=> x = 8/3
1) \(\frac{4x-8}{2x^2+1}=0\)
<=> \(\frac{4\left(x-2\right)}{2x^2+1}=0\)
<=> 4(x - 2) = 0
<=> x - 2 = 0
<=> x = 2
2) \(\frac{x^2-x-6}{x-3}=0\)
<=> \(\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
<=> x + 2 = 0
<=> x = -2
3) xem ở đây Câu hỏi của Vương Thanh Thanh
4) \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
<=> \(\frac{12}{\left(1+3x\right)\left(1-3x\right)}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
<=> 12 = (1 - 3x)2 - (1 + 3x2)
<=> 12 = 1 - 6x + 9x2 - 1 - 6x - 9x2
<=> 12 = -12x
<=> x = -1
5) ĐKXĐ: \(x\ne1,x\ne3\)
\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
<=> \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 5)(x - 3) = (x + 1)(x - 1) - 8
<=> x2 - 3x + 5x - 15 = x2 - x + x - 1 - 8
<=> x2 + 2x - 15 = x2 - 9
<=> x2 + 2x - 15 - x2 = -9
<=> 2x - 15 = -9
<=> 2x = -9 + 15
<=> 2x = 6
<=> x = 3 (ktm)
=> pt vô nghiệm
6) ĐKXĐ: \(x\ne\pm2\)
\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)
<=> \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)
<=> (x + 1)(x + 2) - 5(x - 2) = 12 + (x - 2)(x + 2)
<=> x2 + 2x + x + 2 - 5x + 10 = 12 + x2 + 2x - 2x - 4
<=> x2 - 2x + 12 = x2 + 8
<=> x2 - 2x + 12 - x2 = 8
<=> -2x + 12 = 8
<=> -2x = 8 - 12
<=> -2x = -4
<=> x = 2 (ktm)
=> pt vô nghiệm
\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình trên là \(-2\)