\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}=x-y\) với >0,y>0,x khác y
Những câu hỏi liên quan
CM:
\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{6}}{6}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}+\sqrt{y}}=x-y\) với x.0, y>0, x≠y
\(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{x}}{y-\sqrt{xy}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)với x>0, y>0, x≠y
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
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Rút gọn các biểu thức sau:a) Adfrac{xsqrt{y}+ysqrt{x}}{x+2sqrt{xy}+y}(x≥0 , y≥0 , xy≠0)b) Bdfrac{xsqrt{y}-ysqrt{x}}{x-2sqrt{xy}+y}(x≥0 , y≥0 , x≠y)c) Cdfrac{3sqrt{a}-2a-1}{4a-4sqrt{a}+1}(a≥0 , a≠dfrac{1}{4})d) Ddfrac{a+4sqrt{a}+4}{sqrt{a}+2}+dfrac{4-a}{sqrt{a}-2}(a≥0 , a≠4)
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Rút gọn các biểu thức sau:
a) A=\(\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)(x≥0 , y≥0 , xy≠0)
b) B=\(\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)(x≥0 , y≥0 , x≠y)
c) C=\(\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)(a≥0 , a≠\(\dfrac{1}{4}\))
d) D=\(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)(a≥0 , a≠4)
a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)
\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)
\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)
\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)
\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)
d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
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Chứng minh : A < 0 với y > x > 0
A = \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}=\dfrac{2\sqrt{xy}}{x-y}\)
đk : \(x\ge0;y\ge0;x\ne y\)
A = \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Leftrightarrow\) \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Leftrightarrow\) \(\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Leftrightarrow\) \(\dfrac{x-\sqrt{xy}-\sqrt{xy}-y}{x-y}=\dfrac{2\sqrt{xy}}{x-y}\)
\(\Rightarrow\) \(x-2\sqrt{xy}-y=2\sqrt{xy}\) \(\Leftrightarrow\) \(x-y=4\sqrt{xy}\)
\(\Leftrightarrow\) A = \(\dfrac{2\sqrt{xy}}{4\sqrt{xy}}=\dfrac{1}{2}\)
không biết sai chỗ nào ??? sao bài làm lại trái với câu hỏi thế này ???
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\(\dfrac{\sqrt{y}}{x-\sqrt{ }xy}+\dfrac{\sqrt{x}}{y-\sqrt{ }xy}=-\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
Chứng minh đẳng thức với x ; y lớn hơn 0 ; x khác y
Chứng minh rằng: A = \(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)không phụ thuộc vào x;y với x > 0 và y > 0
Các bạn lm chi tiết giúp mk nhé!
\(A=\dfrac{x-2\sqrt{xy}+y+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\sqrt{x}+\sqrt{y}\\ A=\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}=2\sqrt{y}\)
Đề sai
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\(A=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}+\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)
\(=\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}\)
\(=2\sqrt{x}\)
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Bài 1: Cho Aleft(dfrac{x-y}{sqrt{x}-sqrt{y}}+dfrac{sqrt{x^3}-sqrt{y^3}}{y-x}right):dfrac{left(sqrt{x}-sqrt{y}right)^2+sqrt{xy}}{sqrt{x}+sqrt{y}}với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A dfrac{xsqrt{x}-1}{x-sqrt{x}}-dfrac{xsqrt{x}+1}{x+sqrt{x}}+left(sqrt{x}-dfrac{1}{sqrt{x}}right).left(dfrac{sqrt{x}+1}{sqrt{x}-1}+dfrac{sqrt{x}-1}{sqrt{x}+1}right)
với x0; x≠1
a) Rút gọn A
b)Tìm x để A6
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Bài 1: Cho A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A= \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
với x>0; x≠1
a) Rút gọn A
b)Tìm x để A=6
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
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P=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
1 rút gọn
2 c/m P\(\ge\)0
1. ĐKXĐ : \(xy>0\)
Ta có : \(P=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{-\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-2\sqrt{xy}+y+\sqrt{xy}}\right)\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-2\sqrt{xy}+y+\sqrt{xy}}\right)\)
\(=\dfrac{\left(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-\left(x+\sqrt{xy}+y\right)}{x-\sqrt{xy}+y}=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
2. Ta thấy : \(x-\sqrt{xy}+y=x-\dfrac{2.\sqrt{x}.\sqrt{y}}{2}+\dfrac{y}{4}+\dfrac{3y}{4}\)
\(=\left(\sqrt{x}-\dfrac{\sqrt{y}}{2}\right)^2+\dfrac{3y}{4}\)
Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-\dfrac{\sqrt{y}}{2}\right)^2\ge0\\\dfrac{3y}{4}\ge0\end{matrix}\right.\)
\(\Rightarrow x-\sqrt{xy}+y\ge0\)
Lại có : \(\sqrt{xy}\ge0\)
\(\Rightarrow P\ge0\) ( ĐPCM )
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Giải hệ sau
\(\left\{{}\begin{matrix}\sqrt{\dfrac{x}{y}+\sqrt{\dfrac{y}{x}}}=\dfrac{7}{\sqrt{xy}+1}\\x\sqrt{xy}+y\sqrt{xy}=78\\x>0\\y>0\end{matrix}\right.\)
Cm
a.\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
b. \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{x}}{y-\sqrt{xy}}=-\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)với x>0, y>o và x≠y
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
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