tan2x = cot\(\left(x+\dfrac{\pi}{4}\right)\)

<=>\(\dfrac{1}{cot2x}\)=\(\cot\left(x+\dfrac{\pi}{4}\right)\)

<=>1= \(\cot\left(x+\dfrac{\pi}{4}\right)\) . cot2x

<=>1=\(\dfrac{cos\left(x+\dfrac{\pi}{4}\right)}{sin\left(x+\dfrac{\pi}{4}\right)}\) . \(\dfrac{cos2x}{sin2x}\)

<=>1=\(\dfrac{\dfrac{1}{2}\left[cos\left(x+\dfrac{\pi}{4}-2x\right)+cos\left(x+\dfrac{\pi}{4}+2x\right)\right]}{\dfrac{1}{2}\left[cos\left(x+\dfrac{\pi}{4}-2x\right)-cos\left(x+\dfrac{\pi}{4}+2x\right)\right]}\)

<=>1=\(\dfrac{cos\left(\dfrac{\pi}{4}-x\right)+cos\left(3x+\dfrac{\pi}{4}\right)}{cos\left(\dfrac{\pi}{4}-x\right)-cos\left(3x+\dfrac{\pi}{4}\right)}\)

r b quy đồng x giải pt là ra

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

\(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\) (Đkxđ: \(x\ne-7;x\ne\frac{3}{2}\))

\(\Rightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-9x-4x-6x^2-42x-x=7-6\)

\(\Leftrightarrow-56x=1\)

\(\Leftrightarrow x=-\frac{1}{56}\) (t/m đkxđ)

Vậy \(S=\left\{-\frac{1}{56}\right\}\)

ĐKXĐ: x khác -7 và 3/2

Từ đề bài <=> (3x-2)(2x-3) = (6x+1)(x+7)

<=> 6x^2-4x-9x+6 = 6x^2+x+42x+7

<=> -13x+6 = 43x+7

<=> 6-7 = 43x+13x

<=> 56x = -1

<=> x = -1/56 (TM)

Vậy ...

ĐKXĐ:x khác -7;x khác 1,5

=>(3x-2)(2x-3)=(6x+1)(x+7)

=>6x²-4x-9x+6=6x²+x+42x+7

=>6x²-13x+6=6x²+43x+7

=>6x²-6x²-13x-43x+6-7=0

=>-56x-1=0

=>-56x=1

=>x=\(\frac{-1}{56}\)

Đk \(x+3\ge0\Leftrightarrow x\ge-3\)

Đặt \(\sqrt{\frac{x+3}{2}}=t+1\left(t\ge-1\right)\Leftrightarrow x+3=2\left(t+1\right)^2\Leftrightarrow2t^2+4t=x+1\)

Ta có hệ phương trình:

\(\hept{\begin{cases}2x^2+4x=t+1\\2t^2+4t=x+1\end{cases}}\)

Hệ phương trình  đối xứng loại 2 :). Em làm tiếp nhé:)

ĐKXĐ: \(x>0\)

Ta có:

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

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

\(\frac{\Leftrightarrow1}{2x\sqrt{x}}-\sqrt{x}=2\left(x-\frac{1}{x}+\frac{1}{4x^3}\right)\)

Đặt : \(\frac{1}{2x\sqrt{x}}-\sqrt{x}=a\Rightarrow a^2=x-\frac{1}{x}+\frac{1}{4x^3}\)

Khi đó pt đã cho trở thành:

\(a=2a^2\Leftrightarrow\orbr{\begin{cases}a=0\\a=\frac{1}{2}\end{cases}}\)

+) a = 0\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Tương tự

Theo bài ra , ta có :

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(x\ne3,x\ne-3,x\ne-\frac{7}{2}\)

Quy đồng và khử mẫu phương trình ta đk :

\(13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(13+x-3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+10\right)=12x+42\)

\(\Leftrightarrow x^2+13x+30=12x+42\)

\(\Leftrightarrow x^2+13x-12x+30-42=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2-3x+4x-12=0\)

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

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

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

Kết hợp với ĐKXĐ ta có : x = -4

Vậy \(S=\left\{-4\right\}\)

Chúc bạn học tốt =))

ĐKXĐ: x\(\ne\)3;-7/2;-3

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\Leftrightarrow\frac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\frac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\\ \Leftrightarrow x^2+x=12\)

\(\Leftrightarrow x^2+x-12=0\Leftrightarrow x^2-3x+4x-12=0\\ \Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)

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

Nhận thấy x=3 không thỏa mãn ĐKXĐ nên pt có 1 nghiệm duy nhất là x=-4

Đk: x khác 3; -3 và -7/2

Từ đề suy ra:

13/(x-3)(2x+7) + 1/2x+7 - 6/(x-3)(x+3) = 0

<=> \(\frac{13\left(x+3\right)+\left(x-3\right)\left(x+3\right)-6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=0\)

<=> 13(x+3) + (x-3)(x+3) + 6(2x+7) = 0

<=> 13x + 39 + x² - 9 + 12x + 42 = 0

<=> x² + 25x + 72 = 0

<=> (x + 25/2)² = 337/4

<=> \(\left[\begin{matrix}x+\frac{25}{2}=\sqrt{\frac{337}{4}}\\x+\frac{25}{2}=-\sqrt{\frac{337}{4}}\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=\frac{-25+\sqrt{337}}{2}\\x=\frac{25-\sqrt{337}}{2}\end{matrix}\right.\)(TM)

Vây ...

a) \(\frac{2-x}{3}< \frac{3-2x}{5}\)

<=> \(10-5x< 9-6x\)

<=> x < - 1 

Vậy S = { x| x < -1 }

b)

 

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\frac{1}{x+1+\frac{1}{x}}+\frac{2}{x+2+\frac{1}{x}}=\frac{8}{15}\)

Đặt \(x+1+\frac{1}{x}=a\)

\(\frac{1}{a}+\frac{2}{a+1}=\frac{8}{15}\)

\(\Leftrightarrow a+1+2a=\frac{8}{15}a\left(a+1\right)\)

\(\Leftrightarrow8a^2-37a-15=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-\frac{3}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1+\frac{1}{x}=5\\x+1+\frac{1}{x}=-\frac{3}{8}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+1=0\\x^2+\frac{11}{8}x+1=0\end{matrix}\right.\)