\(A=\sqrt{\frac{25}{5}}+\sqrt{\frac{25}{4}.\frac{4}{5}}-3\sqrt{5}=\sqrt{5}+\sqrt{5}-3\sqrt{5}=-\sqrt{5}\)

\(B=\frac{3}{\sqrt{2}}+\sqrt{\frac{9}{2}}-\sqrt{\frac{25}{2}}=\frac{3}{\sqrt{2}}+\frac{3}{\sqrt{2}}-\frac{5}{\sqrt{2}}=\frac{1}{\sqrt{2}}\)

\(C=\frac{1}{\sqrt{3}}+\sqrt{\frac{6}{5}}+\frac{12}{\sqrt{3}}=\frac{13}{\sqrt{3}}+\sqrt{\frac{6}{5}}\)

\(D=2\sqrt{\frac{16y}{10}}+\sqrt{\frac{225y}{10}}-\sqrt{\frac{100y}{10}}=4\sqrt{\frac{y}{10}}+15\sqrt{\frac{y}{10}}-10\sqrt{\frac{y}{10}}=9\sqrt{\frac{y}{10}}\)

a, Với x > 0 ; \(x\ne1\)

\(M=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)

\(=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{x-1}\right):\left(\frac{2}{x}+\frac{x-2}{x\left(\sqrt{x}+1\right)}\right)\)

\(=\frac{x+2\sqrt{x}}{x-1}:\frac{2\left(\sqrt{x}+1\right)+x-2}{x\sqrt{x}+x}=\frac{x+2\sqrt{x}}{x-1}:\frac{2\sqrt{x}+x}{x\sqrt{x}+x}\)

\(=\frac{x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x}{\sqrt{x}-1}\)

b, Ta có : M = -1/2 => \(\frac{x}{\sqrt{x}-1}=-\frac{1}{2}\Rightarrow2x=-\sqrt{x}+1\)

\(\Leftrightarrow2x+\sqrt{x}-1=0\)Đặt \(\sqrt{x}=t\left(t\ge0\right)\)

\(\Leftrightarrow2t^2+t-1=0\Leftrightarrow\left(2t-1\right)\left(t+1\right)=0\Leftrightarrow t=\frac{1}{2}\left(tm\right);t=-1\left(ktm\right)\)

Theo cách đặt : \(\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)( tmđk )

c, Ta có : \(M>1\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-1}-1>0\Leftrightarrow\frac{1}{\sqrt{x}-1}>0\)

\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)

\(MA^4+MB^4+MC^4+MD^4\)

\(=\left(MA^2+MC^2\right)^2+\left(MB^2+MD^2\right)^2-2MA^2.MC^2-2MB^2.MD^2\)

\(=32R^4-8S_{MAC}^2-8S_{MBD}^2\)

\(=32R^4-8R^2\left(MH^2+MK^2\right)\) với H,K lần lượt là hình chiếu vuông góc của M trên AC,BD

\(=32R^4-8R^2.R^2=24R^4\)

17 nha

\(N=\frac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{y}}\)

\(N=\frac{1+\sqrt{x}+\sqrt{y}\left(\sqrt{x}+1\right)}{1+\sqrt{y}}\)

\(N=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)}{1+\sqrt{y}}\)

\(N=\sqrt{x}+1\)

\(P=\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(P=\frac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(P=\frac{\sqrt{y}\left(4\sqrt{y}+7\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(P=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(P=\sqrt{y}-1\)

\(C=\sqrt{x}+\sqrt{y}+\sqrt{x^2y}+\sqrt{xy^2}\)

\(C=\sqrt{x}\left(\sqrt{xy}+1\right)+\sqrt{y}\left(\sqrt{xy}+1\right)\)

\(C=\left(\sqrt{xy}+1\right)\left(\sqrt{x}+\sqrt{y}\right)\)

\(D=x+2\sqrt{xy}+y-4\)

\(D=\left(\sqrt{x}+\sqrt{y}\right)^2-4\)

\(D=\left(\sqrt{x}+\sqrt{y}-4\right)\left(\sqrt{x}+\sqrt{y}+4\right)\)

\(E=x+\sqrt{x}+\frac{1}{4}-\frac{49}{4}\)

\(E=\left(\sqrt{x}+\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)

\(E=\left(\sqrt{x}+\frac{1}{2}-\frac{7}{2}\right)\left(\sqrt{x}+\frac{1}{2}+\frac{7}{2}\right)\)

\(E=\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)\)

\(F=2a-5\sqrt{ab}+3b\)

\(F=2a-2\sqrt{ab}-3\sqrt{ab}+3b\)

\(F=2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

\(F=\left(2\sqrt{a}-3\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

\(\hept{\begin{cases}\left(x+3\right)\left(y-5\right)=xy\\\left(x-2\right)\left(y+5\right)=xy\end{cases}}\Leftrightarrow\hept{\begin{cases}xy-5x+3y-15=xy\\xy+5x-2y-10=xy\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-5x+3y=15\\5x-2y=10\end{cases}}\Leftrightarrow\hept{\begin{cases}y=25\\5x-2y=10\end{cases}}\Leftrightarrow\hept{\begin{cases}x=12\\y=25\end{cases}}\)

ĐK : \(\hept{\begin{cases}x\ne1\\y\ne-1\end{cases}}\)

\(\hept{\begin{cases}\frac{x+1}{x-1}=\frac{y+3}{y+1}\\3x+2y=0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=\left(x-1\right)\left(y+3\right)\\3x+2y=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}xy+x+y+1=xy+3x-y-3\\3x+2y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-2y=4\\3x+2y=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5x=4\\3x+2y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{4}{5}\\y=-\frac{6}{5}\end{cases}\left(tm\right)}\)

\(cos^4a-sin^4a+2sin^2a\)

\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+2sin^2a\)

\(=cos^2a\left(cos^2a+sin^2a\right)+2sin^2a\)

Bài làm này chắc ổn hơn bài làm trước ✔

\(cos^4a-sin^4a+2sin^2a\)

\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+2sin^2a\)

\(=\left(cos^2a-sin^2a\right)1+2sin^2a\)

\(=cos^2a-sin^2a+2sin^2a\)

\(=cos^2a+sin^2a\)

\(=1\)