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a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)
\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
Còn lại tương tự
1) \(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\8x+3\left(2-4x\right)=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=1\end{matrix}\right.\)
2) 2 pt 3 ẩn không giải được.
3) \(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\3x+2\left(x-2\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+1}{2}\\-4\cdot\frac{3y+1}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}2x+3y=5\\5x-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-3y+5}{2}\\5\cdot\frac{-3y+5}{2}-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}3x-y=7\\x+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2\left(3x-7\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}x+4y=2\\3x+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2-4y\\3\left(2-4y\right)+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{1}{5}\\x=\frac{6}{5}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}-x-y=2\\-2x-3y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-x-2\\-2x-3\left(-x-2\right)=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
Bài 2:
a: 2x+y=1 và x-y=2
=>3x=3 và x-y=2
=>x=1 và y=-1
b: x+2y=2 và x+2y=5
=>0x=-3 và x+2y=2
=>\(\left(x,y\right)\in\varnothing\)
c: 2x+y=3 và -2x-y=-3
=>0x=0 và 2x+y=3
=>\(\left\{{}\begin{matrix}x\in R\\y=3-2x\end{matrix}\right.\)
a, \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
b/
Lần lượt cộng trừ vế cho vế ta được:
\(\left\{{}\begin{matrix}x^3+y^3=7\left(x+y\right)\\x^3-y^3=19\left(x-y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2-xy-7\right)=0\\\left(x-y\right)\left(x^2+y^2+xy-19\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\\\left\{{}\begin{matrix}x^2+y^2-xy-7=0\\x^2+y^2+xy-19=0\end{matrix}\right.\end{matrix}\right.\)
Hai trường hợp đầu bạn tự thế vào giải
Trường hợp 3, trừ vế cho vế: \(2xy-12=0\Rightarrow xy=6\Rightarrow y=\frac{6}{x}\)
Thế vào pt đầu: \(x^3=13x-\frac{36}{x}\Leftrightarrow x^4-13x^2+36=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2=9\\x^2=4\end{matrix}\right.\)
a/ Trừ vế cho vế:
\(2x-2y=y^2-x^2-4y+4x\)
\(\Leftrightarrow x^2-y^2-2x+2y=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=2-x\end{matrix}\right.\)
Thế vào pt đầu:
\(\Rightarrow\left[{}\begin{matrix}2x=x^2-4x+5\\2x=\left(2-x\right)^2-4\left(2-x\right)+5\end{matrix}\right.\)
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