với x>y≥0, biểu thức:\(\dfrac{1}{y-x}\sqrt{x^6\left(x-y\right)2}\)rút gọn bt ta đc...
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Với x > y ≥ 0 , biểu thức: \(\dfrac{1}{y-x}\sqrt{x^6\left(x-y\right)^2}\)có kết quả rút gọn là
\(=\dfrac{1}{y-x}\cdot x^3\cdot\left(x-y\right)=-x^3\)
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Rút gọn biểu thức 1) dfrac{sqrt{14}-sqrt{21}}{sqrt{7}} .2) dfrac{sqrt{a^2+5a+6}}{sqrt{a+3}}3) sqrt{3left(x^2-10x+25right)}.sqrt{27} với x 54)dfrac{y}{x}sqrt{dfrac{x^2}{y^4}} với x 0; y 05) dfrac{1}{x-y}.sqrt{x^6left(x-yright)^4} với x ne y
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Rút gọn biểu thức 1) \(\dfrac{\sqrt{14}-\sqrt{21}}{\sqrt{7}}\) .
2) \(\dfrac{\sqrt{a^2+5a+6}}{\sqrt{a+3}}\)
3) \(\sqrt{3\left(x^2-10x+25\right)}.\sqrt{27}\) với x < 5
4)
\(\dfrac{y}{x}\sqrt{\dfrac{x^2}{y^4}}\) với x > 0; y < 0
5) \(\dfrac{1}{x-y}.\sqrt{x^6\left(x-y\right)^4}\) với x \(\ne\) y
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
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1.giải hệ phương trình:
\(\left\{{}\begin{matrix}2x-y=3\\x+y=0\end{matrix}\right.\)
2.Rút gọn biểu thức
\(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\) với x\(\ge\)0;x\(\ne\)4
1) Ta có: \(\left\{{}\begin{matrix}2x-y=3\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=3\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-x=-1\end{matrix}\right.\)
Vậy: (x,y)=(1;-1)
2) Ta có: \(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\)
\(=\dfrac{x+20+2\left(\sqrt{x}-2\right)-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+20+2\sqrt{x}-4-6\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
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Rút gọn các biểu thức sau:
b) \(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\) x \(\ne\) 1, y \(\ne\) 1, y > 0
a) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)
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b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)
\(=\dfrac{1}{x-1}\)
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Rút gọn biểu thức sau A = \(\left(\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}\right):\left(\sqrt{\dfrac{x}{y}}-\sqrt{\dfrac{y}{x}}\right)\) với x = 1 + a; y = 1 - a
Lời giải:
\(A=\frac{x+y}{\sqrt{xy}}: \frac{x-y}{\sqrt{xy}}=\frac{x+y}{\sqrt{xy}}.\frac{\sqrt{xy}}{x-y}=\frac{x+y}{x-y}\)
\(=\frac{1+a+1-a}{1+a-(1-a)}=\frac{2}{2a}=\frac{1}{a}\)
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Rút gọn biểu thức P=\(\sqrt{\dfrac{1}{x^2+y^2}+\dfrac{1}{\left(x+y\right)^2}+\sqrt{\dfrac{1}{x^4}+\dfrac{1}{y^4}+\dfrac{1}{\left(x^2+y^2\right)^2}}}\)
Mn giúp e với
Rút gọn các biểu thức sau:
a, \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
b, \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) với \(x\ge1\)
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)( do \(x\ge1\))
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a: Ta có: \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
\(=\sqrt{xy}\)
b: Ta có: \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\dfrac{ \left|\sqrt{x}-1\right|}{\left|\sqrt{x}+1\right|}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
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Rút gọn biểu thức
\(a.\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(b.\sqrt{41-\sqrt{160}}+\sqrt{49+\sqrt{90}}\)
\(c.\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne y\right)\)
\(d.\dfrac{y+1-2\sqrt{y}}{\sqrt{y}-1}\left(y\ge0;y\ne1\right)\)
\(e.\sqrt{x+2+2\sqrt{x+1}}-\sqrt{x+2-2\sqrt{x+1}}\)
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: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)
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rút gọn biểu thức: P=\(\dfrac{4\sqrt{xy}}{x-y}\):\(\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y^2}\right)\)-2x
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{\sqrt{y}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}+\sqrt{y}-\sqrt{x}}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-2x\sqrt{x}-2x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
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