Rút gọn biểu thức \(\dfrac{y}{x}\)\(\sqrt[]{\dfrac{x^2}{y^4}}\) (với x >0, y >0) được kết quả là
a. -y
b. \(\dfrac{-1}{y}\)
c. y
d. \(\dfrac{1}{y}\)
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) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
Với x > y ≥ 0 , biểu thức: \(\dfrac{1}{y-x}\sqrt{x^6\left(x-y\right)^2}\)có kết quả rút gọn là
\(=\dfrac{1}{y-x}\cdot x^3\cdot\left(x-y\right)=-x^3\)
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: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
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) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)
b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)
\(=\dfrac{1}{x-1}\)
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: \(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\)
\(=\dfrac{x+20+2\left(\sqrt{x}-2\right)-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+20+2\sqrt{x}-4-6\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
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 biểu thức A
A=(\(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\)) : (\(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\))
với y>0; y≠4; y≠9
\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)
\(A=\dfrac{4\sqrt{y}\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}:\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\)
\(A=\dfrac{8\sqrt{y}+4y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}.\dfrac{\sqrt{y}\left(\sqrt{y}-2\right)}{-\sqrt{y}+3}\)
\(A=\dfrac{4\sqrt{y}}{2-\sqrt{y}}.\dfrac{\sqrt{y}\left(2-\sqrt{y}\right)}{\sqrt{y}-3}\)
\(A=\dfrac{4y}{\sqrt{y}-3}\)
Chúc bạn học tốt ^^
1) Rút gọn biểu thức \(\sqrt{0,81x^2}\) ta được kết quả là ....
2) Rút gọn \(\dfrac{\sqrt{63y^2}}{\sqrt{7y}}\) (với y < 0) ta được kết quả là ....
\(1,=0,9\left|x\right|\\ 2,Sửa:\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3\left|y\right|=-3y\)
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 \(x\ne y,x,y\ge0\))
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{\sqrt{y}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}+\sqrt{y}-\sqrt{x}}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-2x\sqrt{x}-2x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)