a)    \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

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

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

\(\Leftrightarrow\)\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(5x+4\right)=0\)

đến đây tự lm nha

b)   \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)    (1)

ĐKXĐ:    \(x\ne\pm\frac{1}{3}\)

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

\(\Rightarrow\)\(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

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

\(\Leftrightarrow\)\(-12x=12\)

\(\Leftrightarrow\)\(x=-1\)   (t/m ĐKXĐ)

Vậy....

a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

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

\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{4}{5}\end{cases}}}\)

b) ĐKXĐ: \(x\ne\pm\frac{1}{3}\)

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

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

\(\Leftrightarrow\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

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

\(\Leftrightarrow-12x=12\)

\(\Leftrightarrow x=-1\) (thỏa mãn)

Vậy x = -1

a/2(9x²+6x+1)=(3x+1)(x-2)

⇔2(3x+1)²= (3x+1)(x-2)

⇔ 2(3x+1)² :(3x+1)=x-2

⇔ 2(3x+1)=x-2

⇔6x+2-x+2=0

⇔5x+4=0

⇔5x=-4

⇔x=\(\frac{-4}{5}\)

b/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

⇔\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

⇔12=(1-3x)²-(1+3x)²

⇔-(1-3x-1-3x)(1-3x+1+3x)=--12

⇔-(-6x.2)=-12

⇔12x=-12

⇔x=-1

bạn thấy mình làm sai hay thiếu thì bạn nhớ nhắc mình nha.

$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

\(a,\left(4x-7\right)\left(16x^2+28+49\right)=\left(4x\right)^3-7^3=64x^3-343\)\(b,\left(3x+1\right)\left(9x^2-3x+1\right)-9x\left(3x^2+9x\right)\)

\(=27x^3+1-27x^3+9x=9x+1\)

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

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

1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

giúp mình với ạ câu nào cũng được