Lời giải:

$A=2\cos ^3a+\cos ^2a-\sin ^2a+\sin a=2\cos ^3a+2\cos ^2a-1+\sin a$

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

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

$=2(1-\sin a)(1+\sin a)(\cos a+1)-(1-\sin a)$

$=(1-\sin a)[2(\sin a+1)(\cos a+1)-1]$

$=(1-\sin a)(2\sin a\cos a+2\sin a+2\cos a+1)$

$=(1-\sin a)(\sin 2a+2\sin a+2\cos a+1)$

Lời giải:

a) ĐK: $x\geq 0; y\geq 0; x\neq y$

\(A=\left[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$

$\Rightarrow A< 1$

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

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

\(=a-2\sqrt{a}\)

h) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{\sqrt{3-\sqrt{5}}}}+\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{3+\sqrt{5}}}}\)

\(=\sqrt{\dfrac{6+2\sqrt{5}}{\sqrt{2}\left(\sqrt{5}-1\right)}}+\sqrt{\dfrac{6-2\sqrt{5}}{\sqrt{2}\left(\sqrt{5}+1\right)}}\)

\(=\dfrac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\cdot\sqrt{2}}{\sqrt{2}\left(\sqrt{5}-1\right)}+\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\cdot\sqrt{2}}{\sqrt{2}\left(\sqrt{5}+1\right)}\)

\(=\dfrac{4\sqrt{2}}{\sqrt{2}\left(\sqrt{5}-1\right)}+\dfrac{4\sqrt{2}}{\sqrt{2}\left(\sqrt{5}+1\right)}\)

\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)

\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}+3\sqrt{\dfrac{1}{6}}\cdot\sqrt{\dfrac{1}{2}}-\sqrt{12}\)

\(A=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+3\cdot\dfrac{1}{\sqrt{6}}\cdot\dfrac{1}{\sqrt{2}}-2\sqrt{3}\)

\(A=\dfrac{2\sqrt{3}\cdot\left(\sqrt{3}-1\right)}{2}+3\cdot\dfrac{1}{\sqrt{12}}-2\sqrt{3}\)

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

\(A=3-\sqrt{3}+\dfrac{3}{2\sqrt{3}}-2\sqrt{3}\)

\(A=3-3\sqrt{3}+\dfrac{\sqrt{3}}{2}\)

\(A=\dfrac{6+6\sqrt{3}+\sqrt{3}}{2}\)

\(A=\dfrac{6+7\sqrt{3}}{2}\)

Ta có: \(\sqrt{12}-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{11}{2\sqrt{3}+1}\)

\(=2\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{11\left(2\sqrt{3}-1\right)}{\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)}\)

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

\(=\sqrt{3}+2\sqrt{3}-1\)

\(=3\sqrt{3}-1\)

Ta có : \(\sqrt{12}-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{11}{2\sqrt{3}+1}\)

\(=\sqrt{12}-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)}{2\sqrt{3}+1}\)

\(=\sqrt{12}-\sqrt{3}+2\sqrt{3}-1=2\sqrt{3}-\sqrt{3}+2\sqrt{3}-1\)

\(=3\sqrt{3}-1\)

Lời giải:
\(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}=\frac{1+\sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}\)

\(=\frac{1+\sqrt{2}}{1-2}-\frac{\sqrt{2}+\sqrt{3}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}=-(1+\sqrt{2})+(\sqrt{2}+\sqrt{3})-(\sqrt{3}+\sqrt{4})\)

\(=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}=-1-\sqrt{4}=-1-2=-3\)

\(\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+1}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}-\dfrac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{\sqrt{4}+\sqrt{3}}{\left(\sqrt{3}-\sqrt{4}\right)\left(\sqrt{3}+\sqrt{4}\right)}\)

\(=\dfrac{\sqrt{2}+1}{-1}-\dfrac{\sqrt{2}+\sqrt{3}}{-1}+\dfrac{\sqrt{4}+\sqrt{3}}{-1}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{4}-\sqrt{3}\)

\(=-1-\sqrt{4}=-1-2=-3\)

\(\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}\)

\(=-\sqrt{2}-1+\sqrt{3}+\sqrt{2}-2-\sqrt{3}\)

=-3

\(=\dfrac{8+2\sqrt{15}+8-2\sqrt{15}}{2}\)

=8

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