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24 tháng 9 2022
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: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
17 tháng 11 2016
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
17 tháng 11 2016
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
9 tháng 8 2022
a: \(A=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}=10\)
b: \(B=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}=-2\sqrt{y}\)
c: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
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) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
a) đk: \(y\ge0\)
Ta có: \(y-6\sqrt{y}+9=0\)
\(\Leftrightarrow\left(\sqrt{y}-3\right)^2=0\)
\(\Leftrightarrow\sqrt{y}-3=0\)
\(\Leftrightarrow\sqrt{y}=3\)
\(\Rightarrow y=9\left(tm\right)\)
Vậy y = 9
b) đk: \(y\ge2\)
Ta có: \(\sqrt{y^2-4}-3\sqrt{y-2}=0\)
\(\Leftrightarrow\left(\sqrt{y+2}-3\right)\sqrt{y-2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{y+2}=3\\\sqrt{y-2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y+2=9\\y-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=7\left(tm\right)\\y=2\left(tm\right)\end{cases}}\)
Vậy y = 2 hoặc y = 7
a) \(ĐKXĐ:y\ge0\)
\(y-6\sqrt{y}+9=0\)
\(\Leftrightarrow\left(\sqrt{y}-3\right)^2=0\)
\(\Leftrightarrow\sqrt{y}-3=0\)
\(\Leftrightarrow\sqrt{y}=3\)
\(\Leftrightarrow y=9\)( thỏa mãn ĐKXĐ )
Vậy \(x=9\)
b) \(ĐKXĐ:y\ge2\)
\(\sqrt{y^2-4}-3\sqrt{y-2}=0\)
\(\Leftrightarrow\sqrt{\left(y-2\right)\left(y+2\right)}-3\sqrt{y-2}=0\)
\(\Leftrightarrow\sqrt{y-2}.\sqrt{y+2}-3\sqrt{y-2}=0\)
\(\Leftrightarrow\sqrt{y-2}.\left(\sqrt{y+2}-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{y-2}=0\\\sqrt{y+2}-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y-2=0\\\sqrt{y+2}=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=2\\y+2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=2\\y=7\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(y=2\)hoặc \(y=7\)