\(y=\dfrac{x}{\sqrt{a^2-x^2}}\)
(a: const)
K
Khách
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14 tháng 7 2021
a) \(\dfrac{1}{\sqrt{x-1}}=\dfrac{\sqrt{x-1}}{x-1}\)
\(\dfrac{a+2}{\sqrt{a^2-4}}=\dfrac{\sqrt{a+2}}{\sqrt{a-2}}=\dfrac{\sqrt{a^2-4}}{a-2}\)
\(\dfrac{x-y}{\sqrt{x^2-y^2}}=\dfrac{x-y}{\sqrt{\left(x-y\right)\left(x+y\right)}}=\dfrac{\sqrt{x-y}}{\sqrt{x+y}}=\dfrac{\sqrt{x^2-y^2}}{x+y}\)
\(\dfrac{a}{\sqrt{x^2}}=\dfrac{a}{\left|x\right|}\)
b) \(\dfrac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}-1}=\dfrac{\left(\sqrt{x^2-1}+1\right)^2}{x^2-2}\)
c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}=\dfrac{2}{\sqrt{6}-1}=\dfrac{2\left(\sqrt{6}+1\right)}{5}\)
1 tháng 10 2021
1: ĐKXĐ: \(-1< x< 1\)
2: ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)
3: ĐKXĐ: \(\left[{}\begin{matrix}x< -3\\x\ge2\end{matrix}\right.\)
4: ĐKXĐ: \(2< a\le3\)
4 tháng 8 2023
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
B
4 tháng 8 2018
1. \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}+2-\sqrt{a}-2\)
= 0
9 tháng 8 2022
2: \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\dfrac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{y}-\sqrt{x}=0\)
4: \(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\dfrac{1}{1+\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}=\sqrt{a}+1\)
12 tháng 11 2021
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
17 tháng 1 2022
a)\(\sqrt{\frac{3a}{7}}-2\sqrt{\frac{7a}{3}}+\sqrt{21a}\) =\(\sqrt{\frac{3}{7}.\frac{1}{21}.21a}\) - \(2\sqrt{\frac{7}{3}.\frac{1}{21}.21a}\)+ \(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}.21a}\) - \(2\sqrt{\frac{1}{9}.21a}\)+\(\sqrt{21}\)
=\(\sqrt{\frac{1}{49}}.\sqrt{21a}\) - \(2.\sqrt{\frac{1}{9}}.\sqrt{21a}\)+ \(\sqrt{21a}\)
=\(\frac{1}{7}\sqrt{21a}\) - \(\frac{2}{3}\sqrt{21a}\) + \(\sqrt{21a}\)
=\(\frac{-10}{21}\sqrt{21a}\)
17 tháng 1 2022
b)
N=\(\sqrt{\frac{8x}{3}}\) - \(\sqrt{\frac{27x}{2}}\) + \(\sqrt{6x}\)
=\(\sqrt{\frac{8}{3}.\frac{1}{6}.6x}\) - \(\sqrt{\frac{27}{2}.\frac{1}{6}.6x}\)+ \(\sqrt{6x}\)
=\(\frac{2}{3}\sqrt{6x}-\frac{3}{2}.\sqrt{6x}+\sqrt{6x}\)
=\(\frac{1}{6}\sqrt{6x}\)
em lớp 8 nene làm theo cách hiểu thôi ạ
H9
HT.Phong (9A5)
CTVHS
1 tháng 8 2023
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) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
30 tháng 7 2022
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
6 tháng 7 2022
) x y \sqrt{\dfrac{x}{y}}=x y \sqrt{\dfrac{x y}{y^{2}}}=\dfrac{x y}{y} \sqrt{x y}=x \sqrt{x y} .xyyx=xyy2xy=yxyxy=xxy.
b) \dfrac{x}{y} \sqrt{\dfrac{x}{y}}=\dfrac{x}{y} \sqrt{\dfrac{x y}{y^{2}}}=\dfrac{x}{y^{2}} \sqrt{x y} .yxyx=yxy2xy=y2xxy.
c) \sqrt{\dfrac{1}{a}+\dfrac{1}{a^{2}}}=\sqrt{\dfrac{a+1}{a^{2}}}=\dfrac{\sqrt{a+1}}{a} .a1+a21=a2a+1=aa+1.
d) \sqrt{\dfrac{4 x^{3}}{25 y}}=\sqrt{\dfrac{4 x^{2} x y}{25 y^{2}}}=\dfrac{2 x}{5 y} \sqrt{x y} .25y4x3=25y24x2xy=5y2xxy.
e) 2 a b \sqrt{\dfrac{3}{a b}}=2\sqrt{\dfrac{3(ab)^2}{ab}}=2\sqrt{3ab}2abab3=2ab3(ab)2=23ab.
Đề bài yêu cầu điều gì nhỉ?