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đặt \(\dfrac{1}{x+2}=a,\dfrac{1}{y+2}=b\)(\(x,y\ne-2\))
\(=>\left\{{}\begin{matrix}2a+b=1\\8a-5b=1\end{matrix}\right.=>\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{3}\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}\dfrac{1}{x+2}=\dfrac{1}{3}\\\dfrac{1}{y+2}=\dfrac{1}{3}\end{matrix}\right.=>\left\{{}\begin{matrix}x=1\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
a: =>xy-2x+2y-4=xy+y và 5xy+10x+y+2=5xy-10x-2y+4
=>-2x+y=4 và 20x+3y=2
=>x=-5/13; y=42/13
b: =>4x+2|y|=8 và 4x-3y=1
=>2|y|-3y=7 và 4x-3y=1
TH1: y>=0
=>2y-3y=7 và 4x-3y=1
=>-y=7 và 4x-3y=1
=>y=-7(loại)
TH2: y<0
=>-2y-3y=7 và 4x-3y=1
=>y=-7/5; 4x=1+3y=1-21/5=-16/5
=>x=-4/5; y=-7/5
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
Đk: \(y\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=1\\\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)=-1\end{matrix}\right.\)
\(\Rightarrow-\left(x+\dfrac{1}{y}\right)^2+\dfrac{x}{y}=\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)\)
\(\Leftrightarrow-\left(x+\dfrac{1}{y}\right)^2+2\left(x+\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{y}=0\\x+\dfrac{1}{y}=2\end{matrix}\right.\)
TH1: \(x+\dfrac{1}{y}=0\Leftrightarrow\dfrac{1}{y}=-x\) thay vào pt dưới ta được:
\(-x^2=-1\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\Rightarrow y=-1\\x=-1\Rightarrow y=1\end{matrix}\right.\)
TH2: \(x+\dfrac{1}{y}=2\Leftrightarrow\dfrac{1}{y}=2-x\) thay vào pt dưới ta được:
\(\left(2-x\right)x-2.2=-1\)\(\Leftrightarrow x^2-2x+3=0\left(vn\right)\)
Vậy (x;y)=(-1;1);(1;-1)
gợi ý \(\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=1\left(1\right)\\\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)=-1\left(2\right)\end{matrix}\right.\)
Đem \(\left(1\right)+\left(2\right):\left(x+\dfrac{1}{y}\right)^2-2\left(x+\dfrac{1}{y}\right)=0\)
đến đây chắc bạn có thể tự làm được
Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=a\\\dfrac{1}{y+2}=b\end{matrix}\right.\)
\(\Rightarrow\)Ta có hệ mới: \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\cdot\left(3a-2b\right)=2\cdot4\\3\left(2a+b\right)=3\cdot5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6a-4b=8\left(1\right)\\6a+3b=15 \left(2\right)\end{matrix}\right.\)
Lấy (2)-(1) ta đc:
\(\Rightarrow7b=7\Rightarrow b=1\Rightarrow2a+1=5\Rightarrow a=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(x-1\right)\\1=y+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Với \(x\ne1;y\ne-2\)
hpt <=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\dfrac{7x}{x-1}=14\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\2.2+\dfrac{1}{y+2}=5\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}2x-2=x\\\dfrac{1}{y+2}=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=2\\y+2=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
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 \(x=y=z=\dfrac{3}{3}=1\)
\(ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=4\\\dfrac{2}{x}+\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+1=2\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(tm\right)\)
Điều kiện : x ≠ -2 ;y ≠ -2
Đặt : \(\dfrac{1}{x+2}=a;\dfrac{1}{y+2}=b\)
Ta có :
\(hpt\text{⇔}\left\{{}\begin{matrix}2a+b=1\\8x-5b=1\end{matrix}\right.\text{⇔}\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{3}\end{matrix}\right.\)
Suy ra:
\(\left\{{}\begin{matrix}x+2=3\\y+2=3\end{matrix}\right.\text{⇔}\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy nghiệm của hệ phương trình : (x ; y) = (1;1)
Ta có: \(\left\{{}\begin{matrix}\dfrac{2}{x+2}+\dfrac{1}{y+2}=1\\\dfrac{8}{x+2}-\dfrac{5}{y+2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x+2}+\dfrac{4}{y+2}=4\\\dfrac{8}{x+2}-\dfrac{5}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{y+2}=3\\\dfrac{2}{x+2}+\dfrac{1}{y+2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=3\\\dfrac{2}{x+2}=1-\dfrac{1}{3}=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Vậy:(x,y)=(1;1)