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f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
a)\(\frac{3+\sqrt{3}}{1+\sqrt{3}}\)=\(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{1+\sqrt{3}}\)=\(\sqrt{3}\)
b)
\(\frac{y-2\sqrt{y}}{\sqrt{y}-2}\)=\(\frac{\sqrt{y}\left(\sqrt{y}-2\right)}{\sqrt{y}-2}\)=\(\sqrt{y}\)
d) \(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}{\sqrt{x}-1}\)=\(\sqrt{x}\)+3
e)\(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)=\(\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)=\(\sqrt{y}\)-1
g)\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x+3}}\)
chúc bạn học tốt
B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)
=> B = \(\sqrt{2}\)
chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương
\(A=\frac{\sqrt{x}^2+2\sqrt{x}-3}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)}=\sqrt{x}+3\)
\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)
\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)
\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)
\(E=\sqrt{1+2\sqrt{5}+5}+\sqrt{\sqrt{5}-2\sqrt{5}+1}=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
=>\(E=1+\sqrt{5}+\sqrt{5}-1=2\sqrt{5}\)
CÂU CUỐI chưa làm đc
ý cuối cùng này :
\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có
\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)
\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)
\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
a) A=12\(\sqrt{3}\)
B= \(\frac{8}{3}\)
c) C= 1
d)...
Chúc bạn học tốt nha ^^!
ĐKXĐ : \(\left\{{}\begin{matrix}x>7\\y>-6\end{matrix}\right.\)
- Đặt \(\frac{1}{\sqrt{x-7}}=a,\frac{1}{\sqrt{y+6}}=b\) ( \(a,b\ne0\) ) vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}7a-4b=\frac{5}{3}\\5a+3b=\frac{13}{6}\end{matrix}\right.\)
( đoạn này ruễ tự giải nhoa )
=> \(\left\{{}\begin{matrix}a=\frac{1}{3}\\b=\frac{1}{6}\end{matrix}\right.\)( TM )
- Thay lại \(\frac{1}{\sqrt{x-7}}=a,\frac{1}{\sqrt{y+6}}=b\) vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=\frac{1}{3}\\\frac{1}{\sqrt{y+6}}=\frac{1}{6}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-7=9\\y+6=36\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=16\\y=30\end{matrix}\right.\) ( TM )
Vậy .........
