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 .........

THẠNKS

a) Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=a\\\frac{1}{y-1}=b\end{matrix}\right.\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}5a+b=10\\a-3b=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15a+3b=30\\a-3b=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-3b=18\\16a=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=3\\\frac{1}{y-1}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{4}{5}\end{matrix}\right.\)

Vậy...

b) Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=a\\\frac{1}{\sqrt{y+6}}=b\end{matrix}\right.\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\5a+3b=\frac{13}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\51a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{306}\\b=\frac{-43}{612}\end{matrix}\right.\)( loại vì \(a,b>0\) )

Vậy hệ vô nghiệm

Is that true .-.

Cho xin solve lại câu b)

hpt \(\Leftrightarrow\left\{{}\begin{matrix}21a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5a+3b=\frac{13}{6}\\41a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{246}\\b=\frac{8}{123}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=\frac{97}{246}\\\frac{1}{\sqrt{y+6}}=\frac{8}{123}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{126379}{9409}\\y=\frac{14745}{64}\end{matrix}\right.\)

Vậy...

Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)

\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

Trả lời nhanh giúp mình với mình cần gấp lắm

1. 

= -(13 + 3 căn7 ) / 2  +  -(7 + 3 căn7 ) / 2 

=  -7 + 3 căn7