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: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)

\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)

Bài 2.

\(a)5 + \dfrac{{96}}{{{x^2} - 16}} = \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{4 - x}}\)

ĐK: \(x\ne\pm4\)

\( Pt \Leftrightarrow \dfrac{{96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{x - 4}} = - 5\\ \Leftrightarrow \dfrac{{96 - \left( {2x - 1} \right)\left( {x - 4} \right) - \left( {3x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow \dfrac{{ - 5{x^2} - 2x + 96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow - 5{x^2} - 2x + 96 = - 5\left( {{x^2} - 16} \right)\\ \Leftrightarrow 96 - 2x = 80\\ \Leftrightarrow - 2x = - 16\\ \Leftrightarrow x = 8\left( {tm} \right)\\ b)\dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} = \dfrac{{9{x^2}}}{{9{x^2} - 4}} \)

ĐK: \(x \ne \dfrac{2}{3};x \ne -\dfrac{2}{3}\)

\( Pt \Leftrightarrow \dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} - \dfrac{{9{x^2}}}{{9{x^2} - 4}} = 0\\ \Leftrightarrow \dfrac{{{{\left( {2 + 3x} \right)}^2} - 6\left( {3x - 2} \right) - 9{x^2}}}{{\left( {3x - 2} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{16 - 6x}}{{\left( {3 - 2x} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow 16 - 6x = 0\\ \Leftrightarrow - 6x = - 16\\ \Leftrightarrow x = \dfrac{8}{3}\left( {tm} \right)\\ c)\dfrac{{x + 1}}{{{x^2} + x + 1}} - \dfrac{{x - 1}}{{{x^2} - x + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}} \)

Ta có: \(x(x^4+x^2+1)=x[(x^2+1)^2-x^2]=x(x^2+x+1)(x^2-x+1)\)

Do \(\left\{ \begin{array}{l} {x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x\\ {x^2} - x + 1 = \left( {x - \dfrac{1}{2}} \right) + \dfrac{3}{4} > 0\forall x \end{array} \right.\) nên phương trình xác định với mọi $x \ne 0$

Quy đồng, rồi biến đổi phương trình về dạng \(2x=3 \Leftrightarrow x =\dfrac{3}{2} (tm)\)

a) (3x - 2)² - (1 + 5x)²

= (3x - 2 - 1 - 5x)(3x - 2 + 1 + 5x)

= (-2x - 3)(8x - 1)

b) (3x + 4)(3x - 4) - (5 - x)²

= (3x)² - 4² - (25 - 10x + x²)

= 9x² - 16 - 25 + 10x - x²

= 8x²  + 10x - 41

c) \(\left(\dfrac{1}{2}x+4\right)^2-\left(\dfrac{1}{2}x+3\right)\left(\dfrac{1}{2}x-3\right)\)

\(=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.4+4^2-\left[\left(\dfrac{1}{2}x\right)^2-3^2\right]\)

\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)

\(=4x+25\)

a: =9x^2-12x+4-25x^2-10x-1

=-16x^2-22x+3

b: =9x^2-16-x^2+10x-25

=8x^2+10x-41

c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)

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

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

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

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

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

\(\Rightarrow x\left(6x-2-15-6x\right)\)

\(\Rightarrow-16x=0\)

\(\Rightarrow x=0\)

d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)

\(\Rightarrow9x^2-4-4x+4=0\)

\(\Rightarrow9x^2-4x=0\)

\(\Rightarrow x\left(9x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

\(c,2x\left(3x-1\right)-3x\left(5+2x\right)=0\\ \Leftrightarrow6x^2-2x-15x-6x^2=0\\ \Leftrightarrow-17x=0\\ \Leftrightarrow x=0\\ d,\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\\ \Leftrightarrow9x^2-4-4x+4=0\\ \Leftrightarrow9x^2-4x=0\\ \Leftrightarrow x\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(TH_1:3x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(TH_2:-2x-7=0\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

b) \(2x^3-5x^2+3x=0\)

\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)

\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)

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

\(TH_1:x=0\)

\(TH_2:x-1=0\)

\(\Leftrightarrow x=1\)

\(TH_3:2x-3=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)

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

\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)

\(TH_1:3x+4=0\)

\(\Leftrightarrow x=-\dfrac{4}{3}\)

\(TH_2:2x-4=0\)

\(\Leftrightarrow x=2\)

Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)

d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Rightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x-9x=-6-16+12\)

\(\Leftrightarrow11x=-10\)

\(\Leftrightarrow x=-\dfrac{10}{11}\)

Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow3x+1=5x+8\)

\(\Leftrightarrow3x-5x=8-1\)

\(\Leftrightarrow-2x=7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(X=\dfrac{-7}{2}\)

b) Ta có: \(2x^3-5x^2+3x=0\)

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

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

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

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

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

Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow9x^2-16-3x^2-4x=0\)

\(\Leftrightarrow6x^2-4x-16=0\)

\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)

\(\Leftrightarrow3x^2-6x+4x-8=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)

Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)

d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Leftrightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x+16-12-9x+6=0\)

\(\Leftrightarrow11x+10=0\)

\(\Leftrightarrow x=\dfrac{-10}{11}\)

Vậy \(x=\dfrac{-10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

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

\(A=2x^2\left(3x+4\right)\left(3x-4\right)-\dfrac{9}{2}\left(2x^2+1\right)\left(2x^2-1\right)\)

\(=2x^2\left(9x^2-16\right)-\dfrac{9}{2}\left(4x^4-1\right)\)

\(18x^4-32x^2-18x^4+\dfrac{9}{2}\\ =-32x^2+\dfrac{9}{2}\)

Ta có: \(A=2x^2\left(3x+4\right)\left(3x-4\right)-\dfrac{9}{2}\left(2x^2+1\right)\left(2x^2-1\right)\)

\(=2x^2\left(9x^2-16\right)-\dfrac{9}{2}\left(4x^4-1\right)\)

\(=18x^4-36x^2-18x^4+\dfrac{9}{2}\)

\(=-36x^2+\dfrac{9}{2}\)

a)\(\sqrt{3x-1}\)\(=2\)

⇔\(\text{3x-1=2}^2\)

⇔\(3x=5\)

⇔\(x=\dfrac{5}{3}\)

b)\(\sqrt{x^2-4x+4}\)\(\text{=3x-1}\)

⇔\(\text{x-2=3x-1}\)

⇔\(-2x=1\)

⇔\(x=\)\(\dfrac{-1}{2}\)

c)\(\sqrt{x}\)=2-x sai đề bài

d)\(\sqrt{x^2+4}\)=\(\sqrt{3x+8}\)

⇔\(x^2\)\(\text{+4=3x+8}\)

⇔\(x^2\)\(-3x-4=0\)

⇔\(\left(x+1\right)\left(x-4\right)=0\)

⇔\(\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)

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 \(-2\)

\(c.\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\left(x\ne2\right)\\ \Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\\\Leftrightarrow \frac{2\left(x+5\right)}{6\left(x-2\right)}-\frac{3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\\\Leftrightarrow 2\left(x+5\right)-3\left(x-2\right)=3\left(2x-3\right)\\\Leftrightarrow 2x+10-3x+6=6x-9\\\Leftrightarrow 2x-3x-6x=-10-6-9\\\Leftrightarrow -7x=-25\\\Leftrightarrow x=\frac{25}{7}\left(tm\right)\)

Vậy nghiệm của phương trình trên là \(\frac{25}{7}\)