\(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-3}\)  = \(\dfrac{\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{1+2+2\sqrt{2}-3}=\dfrac{1+\sqrt{2}+\sqrt{3}}{2}\)

a) \(x^3-4x^2-8x+8=x^2\left(x+2\right)-6x\left(x+2\right)+4\left(x+2\right)=\left(x^2-6x+4\right)\left(x+2\right)\)

\(=\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)

b) Chưa ra 

c) \(x^2\left(x-2022\right)-x+2022=\left(x^2-1\right)\left(x-2022\right)=\left(x-1\right)\left(x+1\right)\left(x-2022\right)\)

d) \(x^2-8x+15=x\left(x-3\right)-5\left(x-3\right)=\left(x-5\right)\left(x-3\right)\)

e) \(4x^2-3x-1=4x\left(x-1\right)+x-1=\left(4x+1\right)\left(x-1\right)\)

1) ĐK : \(x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

 \(...\Leftrightarrow1+cot^2x=cotx+3\)  \(\Leftrightarrow cot^2x-cotx-2=0\)  

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

Đến đây ; bn làm tiếp

2) \(...\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[sin\left(4x-\dfrac{\pi}{2}\right)+sin2x\right]=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[-cos4x+sin2x\right]=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[2sin^22x-1+sin2x\right]=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(sin^22x+sin2x+1\right)=0\) \(\Leftrightarrow PTVN\)

\(2.5^{x+1}=50\Rightarrow5^{x+1}=5^2\Rightarrow x+1=2\) \(\Rightarrow x=1\)

\(\dfrac{27^{15}}{3^3}=\dfrac{27^{15}}{27}=27^{14}=\left(3^3\right)^{14}=3^{42}\)

sin 2x \(=\dfrac{-\sqrt{2}}{2}\Leftrightarrow sin2x=sin\dfrac{-\pi}{4}\)   \(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{4}+2k\pi\\2x=\dfrac{5\pi}{4}+2k\pi\end{matrix}\right.\)  ( k thuộc Z )

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\)

\(y'=-3x^2+6x-3m=3\left(-x^2+2x-m\right)\)

H/s có 2 điểm cực trị nên : \(y'=0\) có 2 no p/b \(\Leftrightarrow-x^2+2x-m\)  có 2 no p/b \(\Leftrightarrow\Delta'>0\)  \(\Leftrightarrow1-m>0\Leftrightarrow m< 1\)

Viet : \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m\end{matrix}\right.\) 

H/s có 2 điểm cực trị > m nên : \(\left\{{}\begin{matrix}2>2m\\\left[{}\begin{matrix}m>m^2;m>0\\m< m^2;m< 0\end{matrix}\right.\end{matrix}\right.\)  

Sau đấy bn giải ra 

a) \(\left(\dfrac{3}{5}-x\right).\dfrac{7}{11}=-\dfrac{21}{22}\) \(\Rightarrow\dfrac{3}{5}-x=-\dfrac{3}{2}\Rightarrow x=\dfrac{21}{10}\)

b) \(\dfrac{x}{25}=\dfrac{1}{x}\left(x\ne0\right)\Rightarrow x^2=25\Rightarrow x=\pm5\)

c) \(...\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{9}=\dfrac{4}{9}\\x-\dfrac{1}{9}=-\dfrac{4}{9}\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{9}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b. \(B=2\left(x^2-2x+1\right)-1=2\left(x-1\right)^2-1\ge-1\forall x\)  . " = " \(\Leftrightarrow x=1\)

c. Với x \(\ge0\)\(;C=x-6\sqrt{x}+4=\left(\sqrt{x}-3\right)^2-5\ge-5\) . " = " \(\Leftrightarrow x=9\)

d. Với x > 0; \(D=\sqrt{x}+\dfrac{9}{\sqrt{x}}-3\ge2\sqrt{9}-3=3\) ( BĐT Cô-si) 

" = " \(\Leftrightarrow x=9\)