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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: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
\(x^2y+2y+x=4xy< =>xy\left(x+3\right)=4xy< =>x+3=4< =>x=1\)
Thế x=1 vào 1 trong 2 phương trình => y=1
Lời giải:
\(\left\{\begin{matrix} x+\frac{1}{y}=2(1)\\ y+\frac{1}{z}=2(2)\\ z+\frac{1}{x}=2(3)\end{matrix}\right.\)
Lấy \((1)-(2); (2)-(3); (3)-(1)\) ta thu được:
\(\left\{\begin{matrix} x-y+\frac{z-y}{yz}=0\\ y-z+\frac{x-z}{xz}=0\\ z-x+\frac{y-x}{xy}=0\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} x-y=\frac{y-z}{yz}\\ y-z=\frac{z-x}{xz}\\ z-x=\frac{x-y}{xy}\end{matrix}\right.\)
\(\Rightarrow (x-y)(y-z)(z-x)=\frac{(x-y)(y-z)(z-x)}{(xyz)^2}\)
\(\Leftrightarrow (x-y)(y-z)(z-x)(1-\frac{1}{xyz})(1+\frac{1}{xyz})=0\)
TH1: \(x-y=0\Leftrightarrow x=y\Rightarrow x+\frac{1}{x}=2\)
\(\Rightarrow x^2-2x+1=0\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1\rightarrow y=1\)
Thay vào PT\((2)\Rightarrow 1+\frac{1}{z}=2\rightarrow z=1\)
Ta thu được \((x,y,z)=(1,1,1)\)
TH2: \(y-z=0; z-x=0\) hoàn toàn giống TH1 ta cũng có \((x,y,z)=(1,1,1)\)
TH3: \(1-\frac{1}{xyz}=1\Rightarrow xyz=1\)
Thay vào PT(1) và (2)
\(\left\{\begin{matrix} x+\frac{1}{y}=2\\ y+xy=2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} xy+1=2y\\ xy=2-y\end{matrix}\right.\)
\(\Rightarrow 2-y+1=2y\Leftrightarrow y=1\Rightarrow x=z=1\)
TH4: \(1+\frac{1}{xyz}=0\Leftrightarrow xyz=-1\)
Thay vào PT (1) và (2):
\(\left\{\begin{matrix} x+\frac{1}{y}=2\\ y-xy=2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} xy+1=2y\\ xy=y-2\end{matrix}\right.\)
\(\Rightarrow y-2+1=2y\Leftrightarrow y=-1\)
\(\Rightarrow x+\frac{1}{-1}=2\Rightarrow x=3; -1+\frac{1}{z}=2\Rightarrow z=\frac{1}{3}\)
Thử vào PT(3) thấy không đúng (loại)
Vậy \((x,y,z)=(1,1,1)\)
ĐKXĐ : \(x;y\ne0\)
Ta có \(\dfrac{y}{x}-\dfrac{2x}{y}=\dfrac{-5}{2}-\dfrac{2}{xy}\)
\(\Leftrightarrow\dfrac{y^2-2x^2}{xy}=\dfrac{-5xy-4}{2xy}\)
\(\Leftrightarrow2y^2-4x^2+5xy=-4\) (1)
Kết hợp \(x^2+xy-y^2=5\) (2)
ta có : \(-5.\left(2y^2-4x^2+5xy\right)=4\left(x^2+xy-y^2\right)\)
\(\Leftrightarrow16x^2-29xy-6y^2=0\)
\(\Leftrightarrow16x^2-32xy+3xy-6y^2=0\)
\(\Leftrightarrow\left(x-2y\right)\left(16x+3y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=-\dfrac{3y}{16}\end{matrix}\right.\)
Thay \(x=-\dfrac{3y}{16}\) vào (2) ta được
\(\dfrac{9y^2}{256}-\dfrac{3y^2}{16}-y^2=5\)
\(\Leftrightarrow y^2=-\dfrac{256}{59}\Leftrightarrow y\in\varnothing\) (loại)
Khi x = 2y thay vào (2) ta được
4y2 + 2y2 - y2 = 5
\(\Leftrightarrow y=\pm1\) (tm)
Với y = 1 => x = 2
y = -1 => x = -2
Vậy (x;y) = (2;1) ; (-2;-1)
1.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)
Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)
Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)
2.
\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)
\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)
\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)
\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\)
\(\Rightarrow...\)
\(\left\{{}\begin{matrix}x+y+z=1\\\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=64\end{matrix}\right.\)
Ta có:
\(1=x+y+z\ge3\sqrt[3]{xyz}\)
\(\Leftrightarrow xyz\le\dfrac{1}{27}\)
Ta có: \(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=1+\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{xyz}\)
\(\ge1+\dfrac{3}{\sqrt[3]{x^2y^2z^2}}+\dfrac{3}{\sqrt[3]{xyz}}+\dfrac{1}{xyz}\)
\(=1+\dfrac{3}{\sqrt[3]{\dfrac{1}{27^2}}}+\dfrac{3}{\sqrt[3]{\dfrac{1}{27}}}+\dfrac{1}{\dfrac{1}{27}}=64\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{3}\)
Phần cuối là:
\(\ge1+\dfrac{3}{\sqrt[3]{\dfrac{1}{27}}}+\dfrac{3}{\sqrt[3]{\dfrac{1}{27}}}+\dfrac{1}{\dfrac{1}{27}}=64\), không phải là dấu ''=''.
Cái đề có phải là
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=5\\\dfrac{2}{xy}-\dfrac{1}{z^2}=25\end{matrix}\right.????\)