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a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
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\(\left\{{}\begin{matrix}x^2y+xy^2=0\left(1\right)\\2x^2+3xy+2y^2=1\left(2\right)\end{matrix}\right.\)
\(pt\left(1\right)\Leftrightarrow xy\left(x+y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\x=-y\end{matrix}\right.\)
Với \(x=0\) thế vào pt(2) ta được\(2.0^2+3.0.y+2y^2=1\Rightarrow2y^2=1\Rightarrow y^2=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{\sqrt{2}}\)
Với \(y=0\) thế vào pt(2) ta được
\(2x^2+3.x.0+2.0^2=1\Rightarrow2x^2=1\Rightarrow x^2=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{\sqrt{2}}\)
Với \(x=-y\) thế vào pt(2) ta được
\(2\left(-y\right)^2+3\left(-y\right).y+2y^2=1\Rightarrow2y^2-3y^2+2y^2=1\Rightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=1\\y=1\Rightarrow x=-1\end{matrix}\right.\)
vậy ...
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Đặt \(\left\{{}\begin{matrix}x\sqrt{y}=a\\y\sqrt{x}=b\end{matrix}\right.\)
Hpt \(\Leftrightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^2=20\end{matrix}\right.\)
=> Hệ đối xứng loại 1 => EZ
Đặt \(\left\{{}\begin{matrix}a=x\sqrt{y}\\b=\sqrt{x}.y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^2=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^2+\left(6-a\right)^2=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\2a^2-12a+16=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\\left[{}\begin{matrix}a=4\\b=2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\\\left\{{}\begin{matrix}a=2\\b=4\end{matrix}\right.\end{matrix}\right.\)
Trường hợp \(\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{y}=4\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{y}=2\sqrt{x}.y\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{y}-2\sqrt{x}.y=0\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}-2\sqrt{y}\right)=0\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\sqrt{y}\\\sqrt{x}.y=2\end{matrix}\right.\)( vì \(\sqrt{xy}\ne0\) )
\(\Leftrightarrow\left\{{}\begin{matrix}x=4y\\\sqrt{4y}.y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4y\\y\sqrt{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4\end{matrix}\right.\)
TRường hợp 2 tương tự nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}x^2y^2=2x^2+y\\xy^2+2x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2-y=2x^2\\xy^2-1=-2x^2\end{matrix}\right.\)
☘ Cộng vế theo vế
\(\Rightarrow x^2y^2-1+xy^2-y=0\)
\(\Leftrightarrow\left(xy-1\right)\left(xy+1\right)+y\left(xy-1\right)=0\)
\(\Leftrightarrow\left(xy-1\right)\left(xy+1+y\right)=0\)
☘ Trường hợp 1: xy = 1 \(\Leftrightarrow x=\dfrac{1}{y}\)
☘ Trường hợp 2: \(xy+1+y=0\) \(\Leftrightarrow x=-\dfrac{1+y}{y}\)
⚠ Thay vào 1 trong 2 phương trình đề bài cho rồi làm tiếp nhé.
\(x^2y^2+xy+1=x^2\)
\(\Leftrightarrow4x^2y^2+4xy+4=4x^2\)
\(\Leftrightarrow\left(2xy+1\right)^2+3=4x^2\)
\(\Leftrightarrow\left(2x-2xy-1\right)\left(2x+2xy+1\right)=3=1.3=\left(-1\right).\left(-3\right)\)
TH1: \(\left\{{}\begin{matrix}2x-2xy-1=1\\2x+2xy+1=3\end{matrix}\right.\Leftrightarrow...\)
TH2: \(\left\{{}\begin{matrix}2x-2xy-1=3\\2x+2xy+1=1\end{matrix}\right.\Leftrightarrow...\)
TH3: \(\left\{{}\begin{matrix}2x-2xy-1=-1\\2x+2xy+1=-3\end{matrix}\right.\Leftrightarrow...\)
TH4: \(\left\{{}\begin{matrix}2x-2xy-1=-3\\2x+2xy+1=-1\end{matrix}\right.\Leftrightarrow...\)