rút gọn các biểu thức
a) \(\sqrt{\dfrac{x}{y^3}+\dfrac{2x}{y^4}}\)
b) \(\dfrac{x-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c)(a-b)\(\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}\)
d)\(\dfrac{a-\sqrt{3a}+3}{a\sqrt{a}+3\sqrt{3}}\)
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b)\(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{ab}{a-b}=ab\)
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 $x\neq y; x,y\geq 0$
$\Rightarrow A< 1$
a) Sai đề.
\(\dfrac{a+b}{b^2}\sqrt[]{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\left|a\right|}{\left|a+b\right|}=\left|a\right|\)
b) Sai đề.
\(\dfrac{a\sqrt[]{b}+b\sqrt[]{a}}{\sqrt[]{ab}}:\dfrac{1}{\sqrt[]{a}-\sqrt[]{b}}=\dfrac{\sqrt[]{ab}\left(\sqrt[]{a}+\sqrt[]{b}\right)}{\sqrt[]{ab}}.\left(\sqrt[]{a}-\sqrt[]{b}\right)=a-b\)
Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???
b.
\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)
\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)
\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)
a) \(\sqrt{\dfrac{x}{y^3}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy}{y^4}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy+2x}{y^4}}=\dfrac{\sqrt{xy+2x}}{\sqrt{y^4}}=\dfrac{\sqrt{xy+2x}}{\left|y^2\right|}=\dfrac{\sqrt{xy+2x}}{y^2}\)(vì y2\(\ge0\))
b) \(\dfrac{x-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}.\sqrt{x}-\sqrt{x}.\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)
c) \(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right)\dfrac{\sqrt{\left(ab\right)^2}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}\)
Nếu a-b>0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{a-b}=\left|ab\right|\)
Nếu a-b<0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{-\left(a-b\right)}=-\left|ab\right|\)
d) \(\dfrac{a-3\sqrt{a}+3}{a\sqrt{a}+3\sqrt{3}}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}\right)^3+\left(\sqrt{3}\right)^3}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}+\sqrt{3}\right)\left(a-3\sqrt{a}+3\right)}=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)
Nếu trục căn thức ở mẫu thì \(\dfrac{1}{\sqrt{a}+\sqrt{3}}=\dfrac{\sqrt{a}-\sqrt{3}}{\left(\sqrt{a}+\sqrt{3}\right)\left(\sqrt{a}-\sqrt{3}\right)}=\dfrac{\sqrt{a}-\sqrt{3}}{a-3}\)