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\(a.x+y=2\) ⇒ \(\left(x+y\right)^2=4\text{⇒}xy=\dfrac{4-20}{2}=-8\)
Ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(20+8\right)=56\)
\(b.5m-2n=30\text{⇒}\left(5m-2n\right)^2=900\text{⇒}-20mn=900-1200\text{⇒}mn=15\)
a, \(\left\{{}\begin{matrix}3x+5y=3\\5x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+10y=6\\25x+10y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=3\\19x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-3}{19}+5y=3\\x=\dfrac{-1}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{12}{19}\\x=\dfrac{-1}{19}\end{matrix}\right.\)
Vậy hpt có nghiệm (x;y)=(-1/19;12/19)
b, \(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(x-y\right)=5+3x+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\5x-5y-3x-2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\2x-7y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\4x-14y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}23y=-2\\2x-7y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-2}{23}\\x=\dfrac{101}{46}\end{matrix}\right.\)
Vậy hpt có nghiệm (x;y) = ( 101/46;-2/23)
\(\left\{{}\begin{matrix}x+y=3\\x+2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2=3\\y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=3\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}-x+2y=3\\3x+y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}2x+2\sqrt{3}\cdot y=1\\\sqrt{3}x+2y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)