Rút gọn
a)\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)(\(x\ge0;y\ge0;x\ne y\))
b)\(\frac{x\sqrt{x}-8}{x+2\sqrt{x}+4}\)\(\left(x\ge0\right)\)
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10 tháng 8 2021
a: \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+1\)
=2
c: \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
d: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)
NA
14 tháng 10 2018
Q= [\(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\)]\(:\frac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(Q=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(Q=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(Q=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
30 tháng 9 2021
\(a,=\dfrac{x}{y}\cdot\dfrac{\left|x\right|}{y^2}=\dfrac{x^2}{y^3}\\ b,=2y^2\cdot\dfrac{x^2}{\left|2y\right|}=\dfrac{2x^2y^2}{-2y}=-x^2y\)
H9
HT.Phong (9A5)
CTVHS
6 tháng 12 2023
a) \(B=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x,y\ge0;x\ne y\right)\)
\(B=\left[\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{x-y}\right]:\dfrac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{x+\sqrt{xy}+y}\)
b) Xét tử:
\(\sqrt{xy}\ge0\forall x,y\) (xác định) (1)
Xét mẫu:
\(x+\sqrt{xy}+y\)
\(=\left(\sqrt{x}\right)^2+2\cdot\dfrac{1}{2}\sqrt{y}\cdot\sqrt{x}+\left(\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
\(=\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
Mà: \(\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2\ge0\forall x,y\) (xác định), còn: \(\dfrac{3}{4}y\ge0\) vì theo đkxđ thì \(y\ge0\) (2)
Từ (1) và (2) ⇒ B luôn không âm với mọi x,y (\(B\ge0\)) (đpcm)
3 tháng 7 2019
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
3 tháng 7 2019
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
HN
21 tháng 7 2018
a. =\(\frac{x\sqrt{xy}+y\sqrt{x^2}-x\sqrt{y^2}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}\)
=\(\frac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{\sqrt{xy}\left(x-y\right)}{\sqrt{xy}}\)=\(x-y\)
b. =\(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x-1}}\)=\(x+\sqrt{x}+1\)