b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne49\\y>=0\\\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{21}{\sqrt{x}-7}+\dfrac{12}{\sqrt{y}+6}=\dfrac{5}{3}\cdot3=5\\\dfrac{20}{\sqrt{x}-7}+\dfrac{12}{\sqrt{y}+6}=4\cdot\dfrac{13}{6}=\dfrac{52}{6}=\dfrac{26}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{41}{\sqrt{x}-7}=5+\dfrac{26}{3}=\dfrac{41}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=\dfrac{13}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-7=3\\\dfrac{3}{\sqrt{y}+6}=\dfrac{13}{6}-\dfrac{5}{3}=\dfrac{3}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}=10\\\sqrt{y}+6=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=100\left(nhận\right)\\y=0\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: y>=-1
\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(x^2-2x\right)+2\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x^2-2x\right)=-7\\2\left(x^2-2x\right)=-\sqrt{y+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x=-1\\\sqrt{y+1}=-2\left(x^2-2x\right)=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\y+1=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\left(nhận\right)\)
d: \(\left\{{}\begin{matrix}6x+5y=2xy\\30x-20y=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=2xy\\30x-20y=xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+5y=2xy\\60x-40y=2xy\end{matrix}\right.\Leftrightarrow6x+5y=40x-40y\)
=>-34x=-45y
=>\(\dfrac{x}{45}=\dfrac{y}{34}=k\)
=>x=45k; y=34k
6x+5y=2xy
=>\(6\cdot45k+5\cdot34k=2\cdot45k\cdot34k\)
=>\(3060k^2=440k\)
=>\(3060k^2-440k=0\)
=>k(3060k-440)=0
=>\(\left[{}\begin{matrix}k=0\\k=\dfrac{440}{3060}=\dfrac{22}{153}\end{matrix}\right.\)
TH1: k=0
=>\(x=45\cdot0=0;y=34\cdot0=0\)
TH2: \(k=\dfrac{22}{153}\)
=>\(x=45\cdot\dfrac{22}{153}=\dfrac{110}{17};y=34\cdot\dfrac{22}{153}=\dfrac{44}{9}\)
Giusp em với mọi người ạ. Rút gọn câu d, e, f, b giúp em với ạ, Em cảm ơn nhiều lắm.
