bài 4:
a: \(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+18y=16\\-8x+3y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+18y-8x+3y=16+5\\8x+18y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}21x=21\\4x+9y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\9y=8-4x=8-4=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{4}{9}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=\left(x+1\right)\left(y-3\right)\\\left(x-5\right)\left(y+4\right)=\left(x-4\right)\left(y+1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}xy-2x-y+2=xy-3x+y-3\\xy+4x-5y-20=xy+x-4y-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x-y+2=-3x+y-3\\4x-5y-20=x-4y-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x-y+3x-y=-3-2\\4x-5y-x+4y=-4+20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=-5\\3x-y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-6y=-15\\3x-y=16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-6y-3x+y=-15-16\\x-2y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5y=-31\\x=2y-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{31}{5}\\x=2\cdot\dfrac{31}{5}-5=\dfrac{62}{5}-\dfrac{25}{5}=\dfrac{37}{5}\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\left(x-2\right)\left(y+1\right)=xy\\\left(x+8\right)\left(y-2\right)=xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}xy+x-2y-2=xy\\xy-2x+8y-16=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=2\\-2x+8y=16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-4y=4\\-2x+8y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y-2x+8y=4+16\\2x-4y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4y=20\\x=2y+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=2\cdot5+2=10+2=12\end{matrix}\right.\)
Bài 2:
a: \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{4}-2=0\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x+3y}{12}-\dfrac{24}{12}=0\\5x-y=11\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+3y=24\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=24\\15x-3y=33\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=24+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=57\\y=5x-11\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=5\cdot3-11=15-11=4\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{a}{5}+\dfrac{b}{3}=-\dfrac{1}{3}\\4a-5b-10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3a+5b}{15}=\dfrac{-5}{15}\\4a-5b=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3a+5b=-5\\4a-5b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+5b+4a-5b=-5+10\\4a-5b=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7a=5\\5b=4a-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{7}\\5b=4\cdot\dfrac{5}{7}-10=\dfrac{20}{7}-10=-\dfrac{50}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\dfrac{5}{7}\\b=-\dfrac{10}{7}\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\x+y-10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\\dfrac{2}{3}y+y=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10:\dfrac{5}{3}=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)
a) Đặt: `1/x=a;1/y=b` ta có:
\(\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\3\left(b+1\right)+4b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=b+1\\7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\b=\dfrac{2}{7}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{2}{7}+1=\dfrac{9}{7}\\b=\dfrac{2}{7}\end{matrix}\right.=>\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{9}{7}\\\dfrac{1}{y}=\dfrac{2}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y=\dfrac{7}{2}\end{matrix}\right.\)
b) Đặt: `1/x=a;1/y=b` ta có:
\(\left\{{}\begin{matrix}6a+5b=3\\9a-10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12a+10b=6\\9a-10b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}21a=7\\6a+5b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\2+5b=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{5}\end{matrix}\right.=>\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{3}\\\dfrac{1}{y}=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=5\end{matrix}\right.\)
