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1. Đề này là 18 chứ không phải 15 nhé
\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+x+\sqrt{y^2+x+y+1}+y=18\left(1\right)\\\sqrt{x^2+x+y+1}-x+\sqrt{y^2+x+y+1}-y=2\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) và (1) - (2) ta được hệ mới
\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+\sqrt{y^2+x+y+1}=10\\x+y=8\end{matrix}\right.\)
\(\Rightarrow x=8-y\)
\(\Rightarrow\sqrt{x^2+9}+\sqrt{y^2+9}=10\)\(\Leftrightarrow\sqrt{x^2+9}=10-\sqrt{y^2+9}\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2+9=100-20\sqrt{y^2+9}+y^2+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\\left(8-y\right)^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\9y^2-72y+144=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
2. Dễ thấy x = y = 0 không phải là nghiệm của phương trình
HPT\(\Leftrightarrow\left\{{}\begin{matrix}1-\dfrac{12}{y+3x}=\dfrac{2}{\sqrt{x}}\left(1\right)\\1+\dfrac{12}{y+3x}=\dfrac{6}{\sqrt{y}}\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) ; (1) - (2) ta được
\(\left\{{}\begin{matrix}1=\dfrac{1}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}\left(3\right)\\\dfrac{12}{y+3x}=\dfrac{3}{\sqrt{y}}-\dfrac{1}{\sqrt{x}}\left(4\right)\end{matrix}\right.\)
Lấy ( 3) nhân (4)
\(\dfrac{12}{y+3x}=\dfrac{9}{y}-\dfrac{1}{x}=\dfrac{9x-y}{xy}\)
\(\Leftrightarrow27x^2-6xy-y^2=0\Leftrightarrow\left(9x+y\right)\left(3x-y\right)=0\)
\(\Rightarrow y=3x\)
đến đây thì dễ rồi
Điều kiện: \(y\ge0\)
pt thứ nhất của hệ \(\Leftrightarrow\left(y-x+3\right)^2=0\) \(\Leftrightarrow y-x+3=0\) \(\Leftrightarrow y=x-3\)
Thay vào pt thứ hai của hệ, ta được \(2x^2+3x+x-3-\left(3x+1\right)\sqrt{x-3}-2=0\)
\(\Leftrightarrow2x^2+4x-5=\left(3x+1\right)\sqrt{x-3}\) \(\left(x\ge3\right)\)
\(\Rightarrow\left(2x^2+4x-5\right)^2=\left[\left(3x+1\right)\sqrt{x-3}\right]^2\)
\(\Leftrightarrow4x^4+16x^2+25+16x^3-20x^2-40x=\left(3x+1\right)^2\left(x-3\right)\)
\(\Leftrightarrow4x^4+16x^3-4x^2-40x+25=9x^3-21x^2-17x-3\)
\(\Leftrightarrow4x^4+7x^3+17x^2-23x+28=0\)
Đặt \(f\left(x\right)=4x^4+7x^3+17x^2-23x+28\)
\(f\left(x\right)=4x^4+7x^3+17x^2+4+4+...+4-23x+4\) (có 6 số 4 ở giữa)
\(f\left(x\right)\ge9\sqrt[9]{4x^4.7x^3.17x^2.4^6}-23x+4\) \(=\left(9\sqrt[9]{1949696}-23\right)x+4\)
Hiển nhiên \(9\sqrt[9]{1949696}>23\). Lại có \(x\ge3\) nên \(f\left(x\right)>0\), Như vậy pt \(f\left(x\right)=0\) vô nghiệm. Điều đó có nghĩa là phương trình đã cho vô nghiệm.
\(\left\{{}\begin{matrix}x^3-y^3-9=0\\6x^2-12x+3y^2+3y=0\end{matrix}\right.\)
\(\Rightarrow x^3-6x^2+12x-8-\left(y^3+3y^2+3y+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow x-2=y+1\Rightarrow y=x-3\)
Thế vào pt dưới:
\(2x^2+\left(x-3\right)^2-4x+x-3=0\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(x;y\ge1\)
Trừ trên cho dưới:
\(\Rightarrow2\left(\sqrt{x^2+5}-\sqrt{y^2+5}\right)+2\left(\sqrt{x-1}-\sqrt{y-1}\right)+x^2-y^2=0\)
\(\Leftrightarrow\frac{\left(x-y\right)\left(2x+2y\right)}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{2\left(x-y\right)}{\sqrt{x-1}+\sqrt{y-1}}+\left(x-y\right)\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(\frac{2x+2y}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{2}{\sqrt{x-1}+\sqrt{y-1}}+x+y\right)=0\)
\(\Leftrightarrow x-y=0\Rightarrow x=y\)
Thay vào pt đầu:
\(2\sqrt{x^2+5}=2\sqrt{x-1}+x^2\)
\(\Leftrightarrow x^2+2-2\sqrt{x^2+5}+2\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\frac{x^4-16}{x^2+2+2\sqrt{x^2+5}}+\frac{2\left(x-2\right)}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}{x^2+2+2\sqrt{x^2+5}}+\frac{2\left(x-2\right)}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{\left(x+2\right)\left(x^2+4\right)}{x^2+2+2\sqrt{x^2+5}}+\frac{2}{\sqrt{x-1}+1}\right)=0\)
\(\Rightarrow x=y=2\)
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
ĐKXĐ: \(x\le\dfrac{1}{2}\)
\(4x^2+y^2+2x+y=2-4xy\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+2x+y-2=0\)
\(\Leftrightarrow\left(2x+y\right)^2+2x+y-2=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+y=1\\2x+y=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1-2x=y\\1-2x=y+3\end{matrix}\right.\)
Thế vào pt dưới:
\(\Rightarrow\left[{}\begin{matrix}8\sqrt{y}+y^2-9=0\\8\sqrt{y+3}+y^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow...\)