Giải hệ pt : \(x+y+\sqrt{x^2+y^2}=12\)
\(y\sqrt{x^2-y^2}=4\)
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NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 11 2019
Đặt \(\left\{{}\begin{matrix}\sqrt{x+y}=a\ge0\\\sqrt{x-y}=b\ge0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a-b=2\\\sqrt{\frac{a^4+b^4}{2}}+ab=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=2\\\sqrt{\frac{a^4+b^4}{2}}=4-ab\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-b=2\\\frac{a^4+b^4}{2}=a^2b^2-8ab+16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\a^4+b^4-2a^2b^2=16\left(2-ab\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\\left(a^2-b^2\right)^2=16\left(2-ab\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\4\left(a+b\right)^2=16\left(2-ab\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\\left(a+b\right)^2=8-4ab\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=a-2\\\left(2a-2\right)^2=8-4a\left(a-2\right)\end{matrix}\right.\)
\(\Rightarrow4a^2-8a+4=8-4a^2+8a\)
\(\Rightarrow2a^2-4a-1=0\Rightarrow\left[{}\begin{matrix}a=\frac{2+\sqrt{6}}{2}\\a=\frac{2-\sqrt{6}}{2}\left(l\right)\end{matrix}\right.\) \(\Rightarrow b=\frac{\sqrt{6}-2}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+y}=\frac{2+\sqrt{6}}{2}\\\sqrt{x-y}=\frac{\sqrt{6}-2}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=\frac{5+2\sqrt{6}}{2}\\x-y=\frac{5-2\sqrt{6}}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\sqrt{6}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
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:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
4 tháng 3 2020
ĐKXĐ: \(\left\{{}\begin{matrix}2x+y\ge1\\x+2y\ge2\\x+4y\ge0\end{matrix}\right.\)
\(pt\left(1\right)\Leftrightarrow\frac{\left(2x+y-1\right)-\left(x+2y-2\right)}{\sqrt{2x+y-1}+\sqrt{x+2y-2}}+\left(x-y+1\right)=0\)
\(\Leftrightarrow\frac{x-y+1}{\sqrt{2x+y-1}+\sqrt{x+2y-2}}+\left(x-y+1\right)=0\)\(\Leftrightarrow\left(x-y+1\right)\left(\frac{1}{\sqrt{2x+y-1}+\sqrt{x+2y-2}}+1\right)=0\)\(\Leftrightarrow x-y+1=0\)
Thế vào pt 2 => x;y
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 3 2020
Đặt \(\left\{{}\begin{matrix}\sqrt{2x+y-1}=a\ge0\\\sqrt{x+2y-2}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=x-y+1\)
Phương trình thứ nhất trở thành:
\(a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(1+a+b\right)=0\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{2x+y-1}=\sqrt{x+2y-2}\Rightarrow y=x+1\)
Thay xuống pt dưới:
\(4x^2-\left(x+1\right)^2+x+4-\sqrt{3x+1}-\sqrt{5x+4}=0\)
\(\Leftrightarrow3x^2-x+3-\sqrt{3x+1}-\sqrt{5x+4}=0\)
\(\Leftrightarrow3x^2-3x+x+1-\sqrt{3x+1}+x+2-\sqrt{5x+4}=0\)
\(\Leftrightarrow3x\left(x-1\right)+\frac{\left(x+1\right)^2-\left(3x+1\right)}{x+1+\sqrt{3x+1}}+\frac{\left(x+2\right)^2-\left(5x+4\right)}{x+2+\sqrt{5x+4}}=0\)
\(\Leftrightarrow3x\left(x-1\right)+\frac{x\left(x-1\right)}{x+1+\sqrt{3x+1}}+\frac{x\left(x-1\right)}{x+2+\sqrt{5x+4}}=0\)
\(\Leftrightarrow x\left(x-1\right)\left(3+\frac{1}{x+1+\sqrt{3x+1}}+\frac{1}{x+2+\sqrt{5x+4}}\right)=0\)