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a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y\right)^2+\left(x-y\right)^2+\dfrac{3}{\left(x+y\right)^2}=\dfrac{85}{3}\\\left(x+y\right)+\left(x-y\right)+\dfrac{1}{x+y}=\dfrac{13}{3}\end{matrix}\right.\)
\(a=x+y\); \(b=x-y\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a^2+b^2+\dfrac{3}{a^2}=\dfrac{85}{3}\\a+b+\dfrac{1}{a}=\dfrac{13}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(a+\dfrac{1}{a}\right)^2-6+b^2=\dfrac{85}{3}\\a+\dfrac{1}{a}=\dfrac{13}{3}-b\end{matrix}\right.\)
\(\Rightarrow3\left(\dfrac{13}{3}-b\right)^2-6+b^2=\dfrac{85}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}b=1\\b=\dfrac{11}{2}\end{matrix}\right.\)đến đây tự làm nha
1. \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=9\end{matrix}\right.\) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y+xy^2+x+y=5xy\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4y^2+x^2y^4+x^2+y^2=25x^2y^2\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\)\(\Leftrightarrow0=16x^2y^2\)
\(\Rightarrow\) phương trình vô nghiệm
Lời giải:
Lấy PT thứ nhất cộng phương trình thứ 2:
\(\Rightarrow 4(x+y)=\frac{1}{x^2}+\frac{1}{y^2}>0\Rightarrow x+y>0\)
Lấy PT thứ nhất trừ đi phương trình thứ 2:
\((3x+y)-(3y+x)=\frac{1}{x^2}-\frac{1}{y^2}\)
\(\Leftrightarrow 2(x-y)=\frac{y^2-x^2}{x^2y^2}\)
\(\Leftrightarrow (x-y)\left(2+\frac{x+y}{x^2y^2}\right)=0\)
Vì \(x+y>0\Rightarrow 2+\frac{x+y}{x^2y^2}>0\)
Do đó: \(x-y=0\Rightarrow x=y\). Thay vào pt thứ nhất:
\(4x=\frac{1}{x^2}\Rightarrow 4x^3=1\Rightarrow x=\sqrt[3]{\frac{1}{4}}=y\)
5. \(y=\dfrac{-3x}{x+2}\)
xác định khi: \(x+2\ne0\Leftrightarrow x\ne-2\)
vậy D= (\(-\infty;+\infty\))\{-2}
6. \(y=\sqrt{-2x-3}\)
xác định khi: \(-2x-3\ge0\Leftrightarrow x\le\dfrac{-3}{2}\)
vậy D= (\(-\infty;\dfrac{-3}{2}\)]
7. \(y=\dfrac{3-x}{\sqrt{x-4}}\)
xác định khi: x-4 >0 <=> x>4
vậy D= (\(4;+\infty\))
8. \(y=\dfrac{2x-5}{\left(3-x\right)\sqrt{5-x}}\)
xác định khi: \(\left\{{}\begin{matrix}3-x\ne0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x< 5\end{matrix}\right.\)
vậy D= (\(-\infty;5\))\ {3}
9.\(y=\sqrt{2x+1}+\sqrt{4-3x}\)
xác định khi: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{-1}{2}\le x\le\dfrac{4}{3}\)
vậy D= [\(\dfrac{-1}{2};\dfrac{4}{3}\)]
1. \(y=\dfrac{3x-2}{x^2-4x+3}\)
xác định khi : \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)
vậy tập xác định là: D = \(\left(-\infty;+\infty\right)\backslash\left\{3;1\right\}\)
2.\(y=2\sqrt{5-4x}\)
xác định khi \(5-4x\ge0\Leftrightarrow x\le\dfrac{5}{4}\)
vậy D= (\(-\infty;\dfrac{5}{4}\)]
3. \(y=\dfrac{2}{\sqrt{x+3}}+\sqrt{5-2x}\)
xác định khi: \(\left\{{}\begin{matrix}x+3>0\\5-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow-3< x\le\dfrac{5}{2}\)
vậy D= (\(-3;\dfrac{5}{2}\)]
4.\(\sqrt{9-x}+\dfrac{1}{\sqrt{x+2}-2}\)
xác định khi: \(\left\{{}\begin{matrix}9-x\ge0\\x+2\ge0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x\ge-2\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le9\\x\ne2\end{matrix}\right.\)
Vậy D= [\(-2;9\)]\{2}
Điều kiện xác định y>o và x>2
\(\dfrac{5}{x-2}+\dfrac{3}{y}=8\left(1\right)\)
\(\dfrac{2}{x-2}-\dfrac{3}{y}=1\left(2\right)\)
Lấy 1+2 => \(\dfrac{7}{x-2}=9=>7=9.\left(x-2\right)=>x=\dfrac{25}{9}\)(Tm)
Thay x=\(\dfrac{25}{9}\) vào 1 hoặc 2 => \(\dfrac{5}{\dfrac{25}{9}-2}+\dfrac{3}{y}=8=>y=\dfrac{21}{11}\)(TM)
Vậy.........