\(\left(x^2-3x+2\right)\sqrt{\frac{x+3}{x-1}}=-\frac{x^3}{2}+\frac{15x}{2}-11\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\sqrt{\frac{x+3}{x-1}}=-\frac{1}{2}\left(x-2\right)\left(x^2+2x-11\right)\)

\(\Leftrightarrow\left(x-2\right)\left[2\left(x-1\right)\sqrt{\frac{x+3}{x-1}}+\left(x^2+2x-11\right)\right]=0\)

Làm nốt

ĐKXĐ: \(x\le-3\)hoặc 1 < x

(x² - 3x +2)\(\sqrt{\frac{x+3}{x-1}}\)=\(\frac{-1}{2}x^3+\frac{15}{2}x-11\)

<=> (x - 1)(x - 2)\(\sqrt{\frac{x+3}{x-1}}\)=\(\frac{-1}{2}\left(x-2\right)\left(x^2+2x-11\right)\) (1)

+ TH₁: x = 2 là nghiệm của phương trình (1).

+ TH₂: \(x\ne2\). Lấy 2 vế của phương trình (1) chia cho (x - 2), ta được:

(x - 1)\(\sqrt{\frac{x+3}{x-1}}\)=\(\frac{-1}{2}\left(x^2+2x-11\right)\)

Đến đây bạn tự giải tiếp.

a) ĐKXĐ: \(x\notin\left\{\frac{1}{3};\frac{-11}{3}\right\}\)

Ta có: \(\frac{2}{\left(1-3x\right)\left(3x+11\right)}=\frac{1}{9x^2-6x+1}-\frac{3}{\left(3x+11\right)^2}\)

\(\Leftrightarrow\frac{2\left(1-3x\right)\left(3x+11\right)}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}=\frac{\left(3x+11\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}-\frac{3\left(1-3x\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}\)

\(\Leftrightarrow-18x^2-60x+22=9x^2+66x+121-3\left(1-6x+9x^2\right)\)

\(\Leftrightarrow-18x^2-60x+22-9x^2-66x-121+3\left(1-6x+9x^2\right)=0\)

\(\Leftrightarrow-27x^2-126x-99+3-18x+27x^2=0\)

\(\Leftrightarrow-144x-96=0\)

\(\Leftrightarrow-144x=96\)

hay \(x=\frac{-2}{3}\)(tm)

Vậy: \(x=\frac{-2}{3}\)

a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi  - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)

b) \(2\cos x =  - \sqrt 2 \;\; \Leftrightarrow \cos x =  - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x =  - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)

c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)

\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)

a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)

\(=>\frac{1-x+x+1}{x+1}+2=\frac{1}{x+1}+2\)

\(=>\frac{2}{x+1}=\frac{1}{x+1}\)

\(=>2x+2=x+1\)

\(=>2x-x=1-2=-1\)

\(=>x=-1\)

vậy nghiệm của phương trình trên là {-1}

À quên ĐKXĐ của câu a là \(x\ne-1\)

Nên \(x\in\varnothing\)nhé :v

\(\left(2x^2+1\right)\left(4x-2\right)=\left(2x^2+1\right)\left(x-12\right)\)

\(\Leftrightarrow8x^3-6x^2+4x-3=2x^3-24x^2+x-12\)

\(\Leftrightarrow8x^3-6x^2+4x-3-2x^3+24x^4-x+12=0\)

\(\Leftrightarrow6x^3+18x^2+3x+9=0\)

\(\Leftrightarrow3\left(x+3\right)\left(2x^2+1\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)