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a.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2-xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)
\(\Rightarrow x^2+\left(\dfrac{2}{x}\right)^2=5\)
\(\Leftrightarrow x^4-5x^2=4=0\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}\right)^2-\left(y+\dfrac{1}{y}\right)^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)\left(x+\dfrac{1}{x}-y-\dfrac{1}{y}\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\x+\dfrac{1}{x}-y-\dfrac{1}{y}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=5\\y+\dfrac{1}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+1=0\\y^2-2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
b) Áp dụng bđt Svac-xơ:
\(\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}\ge\dfrac{\left(1+3+4\right)^2}{x+y+z}\ge\dfrac{64}{4}=16>9\)
=> hpt vô nghiệm
c) Ở đây x,y,z là các số thực dương
Áp dụng cosi: \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=3xyz\)
Dấu = xảy ra khi \(x=y=z=\dfrac{3}{3}=1\)
a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c: ĐKXĐ: y>2
\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)
câu a rút x ở pt 1 thay vào pt 2
-> đcược pt 4 ẩn tự giải đc
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)=xy+100\\\left(x-2\right)\left(y-2\right)=xy-64\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=94\\-2x-2y=-68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-3x+2y=0\\-x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}xy-2x=xy-4x+2y-8\\2xy+7x-6y-21=2xy+6x-7y-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=-8\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
a)
\(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)
Đặt x+y = S, xy = P,ta có hệ
\(\left\{{}\begin{matrix}S+P=17\\S^2-P=13\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}P=S-17\\S^2-S+4=0\end{matrix}\right.\)
\(S^2-S+4>0\)
=> Hệ phương trình vô nghiệm