rút gọn \(\sqrt{2.\left(\sqrt{x^2+y^2}-x\right)\left(\sqrt{x^2+y^2-y}\right)}+\sqrt{x^2+y^2}\)
Những câu hỏi liên quan
Cho x,y >0. Rút gọn A=\(\sqrt{2\left(\sqrt{x^2+y^2}+x\right)\left(\sqrt{x^2+y^2}+y\right)}-\sqrt{x^2+y^2}-x-y+2020\)
\(A=\sqrt{2\left(x^2+y^2+\left(x+y\right)\sqrt{x^2+y^2}+xy\right)}-\sqrt{x^2+y^2}-x-y+2020\)
\(=\sqrt{\left(x^2+y^2+2xy\right)+x^2+y^2+2\left(x+y\right)\sqrt{x^2+y^2}}-\sqrt{x^2+y^2}-x-y+2020\)
\(=\sqrt{\left(x+y\right)^2+2\left(x+y\right)\sqrt{x^2+y^2}+x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020\)
\(=\sqrt{\left(x+y+\sqrt{x^2+y^2}\right)^2}-\sqrt{x^2+y^2}-x-y+2020\)
\(=x+y+\sqrt{x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020\)
\(=2020\)
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a) \(Q=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+2x\sqrt{x}+y\sqrt{y}}{x\sqrt{x}+y\sqrt{y}}\left(x>0,y>0\right)\)
Rút Gọn
b) \(M=\frac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right)x^2-\sqrt{6}}\)
Rút Gọn
\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^2}-\sqrt{y^2}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a) Rút gọn A
b) Chứng minh A ≥0
a:
Sửa đề: \(A=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(A=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: căn xy>0
\(x-\sqrt{xy}+y=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}\sqrt{y}+\dfrac{1}{4}y+\dfrac{3}{4}y\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y>0\)
=>A>0
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Rút gọn các biểu thức sau:a) A left(dfrac{sqrt{x}}{x-4}+dfrac{2}{2-sqrt{x}}+dfrac{1}{sqrt{x}+2}right):left(sqrt{x}-2+dfrac{10-x}{sqrt{x}+2}right)b) B left(dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-sqrt{xy}right):left(x-yright)+dfrac{2sqrt{y}}{sqrt{x}+sqrt{y}}c) C left(1-dfrac{sqrt{x}}{1+sqrt{x}}right):left(dfrac{sqrt{x}+3}{sqrt{x}-2}+dfrac{2+sqrt{x}}{3-sqrt{x}}+dfrac{sqrt{x}+2}{x-5sqrt{x}+6}right)d) D sqrt{dfrac{a+x^2}{x}-2sqrt{a}}-sqrt{dfrac{a+x^2}{x}+2sqrt{a}} với a 0, x 0.
Đọc tiếp
Rút gọn các biểu thức sau:
a) A = \(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
b) B = \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
c) C = \(\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2+\sqrt{x}}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
d) D = \(\sqrt{\dfrac{a+x^2}{x}-2\sqrt{a}}-\sqrt{\dfrac{a+x^2}{x}+2\sqrt{a}}\) với a > 0, x > 0.
Rút gọn
\(\sqrt{2\left(\sqrt{x^2+y^2}-x\right)\left(\sqrt{x^2+y^2}-y\right)}+\sqrt{x^2+y^2}\)
SGK Kết nối tri thức và cuộc sống
18 tháng 8 2023 lúc 17:58
Rút gọn biểu thức: \({\left( {{x^{\sqrt 2 }}y} \right)^{\sqrt 2 }}\left( {9{y^{ - \sqrt 2 }}} \right)\) (với \(x,y > 0\)).
$\left(x^{\sqrt{2}}y\right)^{\sqrt{2}} = x^{\sqrt{2} \cdot \sqrt{2}}y^{\sqrt{2}} = x^2y^{\sqrt{2}}$
$x^2y^{\sqrt{2}} \cdot 9y^{-\sqrt{2}} = 9x^2y^{\sqrt{2}}y^{-\sqrt{2}} = 9x^2$
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rút gọn Bt
a)\(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
b)\(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(x\ne1,y\ne1,y>0\right)\)
a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=3\sqrt{xy}\)
b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)
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a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)
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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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1) Rút gọn biểu thứ
A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a) Rút gọn A
b) Chứng minh 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 $x\neq y; x,y\geq 0$
$\Rightarrow A< 1$
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Rút gọn:
\(A=\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right]\)
\(x=\sqrt{2-\sqrt{3}};y=\sqrt{2+\sqrt{3}}\)