giúp mình với 

ĐKXĐ: \(x\ne\dfrac{3a}{2}\)

- Với \(a=0\) pt vô nghiệm

- Với \(a\ne0\)

\(\dfrac{2a}{2x-3a}=1\Leftrightarrow2a=2x-3a\)

\(\Leftrightarrow2x=5a\Rightarrow x=\dfrac{5a}{2}\)

a, \(\dfrac{1}{2}\sqrt{x-5}-\sqrt{4x-20+3}=0\left(dkxd:x\ge5\right)\)

\(< =>\dfrac{\sqrt{x-5}}{2}=\sqrt{4x-17}\)

\(< =>\dfrac{x-5}{4}=4x-17\)

\(< =>x-5=16x-68\)

\(< =>15x=68-5=63\)

\(< =>x=\dfrac{63}{15}=\dfrac{21}{5}\)(ktm)

b, \(\sqrt{2x+1}-2\sqrt{x}+1=0\left(dkxd:x\ge0\right)\)

\(< =>\sqrt{2x+1}+1=2\sqrt{x}\)

\(< =>2x+1+1+2\sqrt{2x+1}=4x\)

\(< =>2x-2\sqrt{2x+1}-2=0\)

\(< =>2x+1-2\sqrt{2x+1}+1-4=0\)

\(< =>\left(\sqrt{2x+1}-1\right)^2=4\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}-1=2\\\sqrt{2x+1}-1=-2\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{2x+1}=-1\left(loai\right)\end{matrix}\right.\)

\(< =>2x+1=9< =>2x=8< =>x=4\)(tmdk)

a) \(\sin 2x + 1 - 2{\sin ^2}2x = 0\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin 2x = 1}\\{\sin 2x =  - \frac{1}{2}}\end{array}\;\;\;} \right. \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{\sin 2x = \sin \frac{\pi }{2}}\\{\sin 2x = \sin  - \frac{\pi }{6}}\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = \frac{\pi }{2} + k2\pi }\\{2x =  - \frac{\pi }{6} + k2\pi }\\{2x = \pi  + \frac{\pi }{6} + k2\pi }\end{array}} \right.\;\;\)

\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{4} + k2\pi }\\{x =  - \frac{\pi }{{12}} + k\pi }\\{x = \frac{{7\pi }}{{12}} + k\pi }\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

b) \(\cos 3x =  - \cos 7x\; \Leftrightarrow \cos 3x + \cos 7x = 0\;\; \Leftrightarrow 2\cos 5x\cos 2x = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos 5x = 0}\\{\cos 2x = 0\;}\end{array}} \right.\;\;\)

\( \Leftrightarrow \left[ \begin{array}{l}\cos 5x = \cos \frac{\pi }{2}\\\cos 2x = \cos \frac{\pi }{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\5x =  - \frac{\pi }{2} + k2\pi \\2x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x =  - \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = \frac{\pi }{4} + k\pi \\x =  - \frac{\pi }{4} + k\pi \end{array} \right.;k \in Z\)

a, \(\dfrac{\left(2x-5\right)\left(x+2\right)}{4x-3}< 0\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)< 0\\4x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)>0\\4x-3< 0\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-2< x< \dfrac{5}{2}\\x>\dfrac{3}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -2\\x>\dfrac{5}{2}\end{matrix}\right.\\x< \dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\dfrac{3}{4}< x< \dfrac{5}{2}\\x< -2\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là

S = \(\left(\dfrac{3}{4};\dfrac{5}{2}\right)\cup\left(-\infty;-2\right)\)

b, Pt

⇔ \(\left\{{}\begin{matrix}x^2-5x+6=x^2+6x+5\\x\in R\backslash\left\{-1;2\right\}\end{matrix}\right.\)

⇔ x = \(\dfrac{1}{11}\)

Vậy S = \(\left\{\dfrac{1}{11}\right\}\)

a) \(\sqrt 3 \tan 2x =  - 1\;\; \Leftrightarrow \tan 2x =  - \frac{1}{{\sqrt 3 }}\;\;\; \Leftrightarrow \tan 2x = \tan  - \frac{\pi }{6}\; \Leftrightarrow 2x =  - \frac{\pi }{6} + k\pi \)

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

b) \(\tan 3x + \tan 5x = 0\;\; \Leftrightarrow \tan 3x = \tan \left( { - 5x} \right) \Leftrightarrow 3x =  - 5x + k\pi \;\; \Leftrightarrow 8x = k\pi \;\; \Leftrightarrow x = \frac{{k\pi }}{8}\;\left( {k \in \mathbb{Z}} \right)\)

\(a,\Leftrightarrow\left\{{}\begin{matrix}6x-9y=-15\\-6x+8y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-5\\-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5+33}{2}=14\\y=11\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-3y=-5\\-3x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-9y=-15\\-6x+8y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-y=-11\\2x-3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=11\\x=\dfrac{-5+3y}{2}=\dfrac{-5+3\cdot11}{2}=14\end{matrix}\right.\)

câu 1 

a) 5x(x-2)=0 =>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b)(x+5)(2x-7)=0 =>\(\left[{}\begin{matrix}x+5=0\\2x-7=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-5\\x=\dfrac{7}{2}\end{matrix}\right.\)

a) \(0,{1^{2x - 1}} \le 0,{1^{2 - x}} \Leftrightarrow 2x - 1 \ge 2 - x \Leftrightarrow 3x \ge 3 \Leftrightarrow x \ge 1\)

b) \({3.2^{x + 1}} \le 1 \Leftrightarrow {2^{x + 1}} \le \frac{1}{3} \Leftrightarrow x + 1 \le {\log _2}\frac{1}{3} \Leftrightarrow x \le  - {\log _2}3 - 1 =  - {\log _2}3 - {\log _2}2 =  - {\log _2}6\)

a: \(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)+\sqrt{3}=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)=-\sqrt{3}\)

=>\(sin\left(x+\dfrac{\Omega}{5}\right)=-\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{5}=-\dfrac{\Omega}{3}+k2\Omega\\x+\dfrac{\Omega}{5}=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{8}{15}\Omega+k2\Omega\\x=\dfrac{4}{3}\Omega-\dfrac{\Omega}{5}+k2\Omega=\dfrac{17}{15}\Omega+k2\Omega\end{matrix}\right.\)

b: \(sin\left(2x-50^0\right)=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}2x-50^0=60^0+k\cdot360^0\\2x-50^0=300^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=110^0+k\cdot360^0\\2x=350^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=55^0+k\cdot180^0\\x=175^0+k\cdot180^0\end{matrix}\right.\)

c: \(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)-1=0\)

=>\(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)=1\)

=>\(tan\left(2x-\dfrac{\Omega}{3}\right)=\dfrac{1}{\sqrt{3}}\)

=>\(2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\)

=>\(2x=\dfrac{1}{2}\Omega+k2\Omega\)

=>\(x=\dfrac{1}{4}\Omega+k\Omega\)

a) Ta có \(a = 3 > 0\) và tam thức bậc hai \(f\left( x \right) = 3{x^2} - 2x + 4\) có \(\Delta ' = {1^2} - 3.4 =  - 11 < 0\)

=> \(f\left( x \right) = 3{x^2} - 2x + 4\) vô nghiệm.

=> \(3{x^2} - 2x + 4 > 0\forall x \in \mathbb{R}\)

b) Ta có: \(a =  - 1 < 0\) và \(\Delta ' = {3^2} - \left( { - 1} \right).\left( { - 9} \right) = 0\)

=> \(f\left( x \right) =  - {x^2} + 6x - 9\) có nghiệm duy nhất \(x = 3\).

=> \( - {x^2} + 6x - 9 < 0\forall x \in \mathbb{R}\backslash \left\{ 3 \right\}\)