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a) \(\hept{\begin{cases}\left(x-1\right)\left(2x+y\right)=0\\\left(y+1\right)\left(2y-x\right)=0\end{cases}}\)
\(\cdot x=1\Rightarrow\hept{\begin{cases}0=0\\\left(y+1\right)\left(2y-1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\y=-1;y=\frac{1}{2}\end{cases}}\)
\(\cdot y=-1\Rightarrow\hept{\begin{cases}\left(x-1\right)\left(2x-1\right)=0\\0=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1;x=\frac{1}{2}\\0=0\end{cases}}\)
\(\cdot x=2y\Rightarrow\hept{\begin{cases}\left(2y-1\right)5y=0\\0=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=0\\y=\frac{1}{2}\Rightarrow x=1\end{cases}}\)
\(y=-2x\Rightarrow\hept{\begin{cases}0=0\\\left(1-2x\right)5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\Rightarrow y=-1\\x=0\Rightarrow y=0\end{cases}}\)
b) \(\hept{\begin{cases}x+y=\frac{21}{8}\\\frac{x}{y}+\frac{y}{x}=\frac{37}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\\left(\frac{21}{8}-y\right)^2+y^2=\frac{37}{6}y\left(\frac{21}{8}-y\right)\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\2y^2-\frac{21}{4}y+\frac{441}{64}=-\frac{37}{6}y^2+\frac{259}{16}y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\1568y^2-4116y+1323=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{8}\\y=\frac{9}{4}\end{cases}}hay\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{3}{8}\end{cases}}\)
c) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x-y\right)^2=-4x^2y^2+2xy\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}8x^2y^2-4x^2y-4xy^2+x^2+y^2-2xy+2xy=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4x^2y^2-4x^2y+x^2+4x^2y^2-4xy^2+y^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x\right)^2+\left(2xy-y\right)^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=\frac{-1}{2}\end{cases}}\)
d) \(\hept{\begin{cases}xy+x+y=71\\x^2y+xy^2=880\end{cases}}\). Đặt \(\hept{\begin{cases}x+y=S\\xy=P\end{cases}}\), ta có: \(\hept{\begin{cases}S+P=71\\SP=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P\left(71-P\right)=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P^2-71P+880=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=16\\P=55\end{cases}}hay\hept{\begin{cases}S=55\\P=16\end{cases}}\)
\(\cdot\hept{\begin{cases}S=16\\P=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=16\\xy=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y\left(16-y\right)=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y^2-16y+55=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=11\end{cases}}hay\hept{\begin{cases}x=11\\y=5\end{cases}}\)
\(\cdot\hept{\begin{cases}S=55\\P=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=55\\xy=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y\left(55-y\right)=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y^2-55y+16=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{55-3\sqrt{329}}{2}\\y=\frac{55+3\sqrt{329}}{2}\end{cases}}hay\hept{\begin{cases}x=\frac{55+3\sqrt{329}}{2}\\y=\frac{55-3\sqrt{329}}{2}\end{cases}}\)
e) \(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{cases}}\). Đặt \(\hept{\begin{cases}S=\sqrt{x}+\sqrt{y}\\P=\sqrt{xy}\end{cases}}\), ta có \(\hept{\begin{cases}SP=12\\P\left(S^2-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\P\left(\frac{144}{P^2}-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\2P^4+28P^2-144P=0\end{cases}}\)
Tự làm tiếp nhá! Đuối lắm luôn
Ôi trời nhiều thía ? làm từng câu một ha !
a \(\hept{\begin{cases}\left(x+5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}xy-2x+5y-10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-x+3y=8\\3x-y=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y-3x+9y=16+24\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\8y=40\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=7\\y=5\end{cases}}\)
b, ĐKXĐ \(x\ne\pm y\)
Đặt \(\frac{1}{x+y}=a\) và \(\frac{1}{x-y}=b\)(a và b khác 0)
Ta có hệ \(\hept{\begin{cases}a-2b=2\\5a-4b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b-2a+4b=3-4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\3a=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=-\frac{1}{3}\\b=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+y}=-\frac{1}{3}\\\frac{1}{x-y}=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y=-3\\x-y=-\frac{6}{7}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y-x+y=-3+\frac{6}{7}\\x-y=-\frac{6}{7}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2y=-\frac{15}{7}\\x-y=-\frac{6}{7}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-\frac{27}{14}\\y=-\frac{15}{14}\end{cases}}\)
1.
\(ĐK:x\ne0\)
HPT
\(\Leftrightarrow\hept{\begin{cases}2x\left(x+y\right)-3x+1=0\\3x\left(x+y\right)-x-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x\left(x+y\right)-\frac{9}{2}x+\frac{3}{2}=0\left(1\right)\\3x\left(x+y\right)-x-2=0\left(2\right)\end{cases}}\)
\(\left(1\right)-\left(2\right)\Leftrightarrow\frac{7}{2}x=\frac{7}{2}\)
\(\Leftrightarrow x=1\left(3\right)\)
\(\left(1\right),\left(3\right)\Rightarrow3\left(1+y\right)-3=0\)
\(\Leftrightarrow y=0\)
Vay nghiem cua HPT la \(\left(1;0\right)\)
a) \(\Leftrightarrow\hept{\begin{cases}\frac{x+1+1}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}\Leftrightarrow\hept{\begin{cases}1+\frac{1}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}}\)
Đặt \(a=\frac{1}{x+1};b=\frac{1}{y-2}\)
\(\Leftrightarrow\hept{\begin{cases}1+a+2b=6\\5a-b=3\end{cases}\Leftrightarrow\hept{\begin{cases}a+2b=5\\5a-b=3\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+1}=1\\\frac{1}{y-2}=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=\frac{5}{2}\end{cases}}}\)
b) ĐK: \(\hept{\begin{cases}x\ne0\\y\ne1\end{cases}}\)
\(PT\left(1\right)\Leftrightarrow\left(x^2-2x\right)\left(x^2-2x+4\right)=0\Leftrightarrow x\left(x-2\right)\left(x^2-2x+4\right)=0\Leftrightarrow x=0\)(loại)
, x=2 , x2-2x+4=0 (3)
pt(3) vô nghiệm vì \(\Delta'=1-4=-3< 0\)
Thay x=2 vào pt(2) ta được \(\frac{1}{2}+\frac{1}{y-2}=\frac{3}{2}\Leftrightarrow\frac{1}{y-1}=1\Leftrightarrow y-1=1\Leftrightarrow y=2\left(tm\text{đ}k\right)\)
Vậy nghiệm của hpt là: (x;y)=(2;2)
pt (2)<=>\(x^2y^2\left(x+y\right)=xy\left(x^2+y^2\right)\)
<=>\(xy\left(x^2y+xy^2-x^2-y^2\right)=0\)
TH1 :xy=0
tự giải
TH2
\(x^2y+xy^2-x^2-y^2=0\)
<=>\(xy\left(x+y\right)=x^2+y^2\)
<=>\(x+y=\frac{x}{y}+\frac{y}{x}>=2\)
dau =xay ra <=>x=y
vt cua (1): \(\sqrt{x+y+2}+x+y=\sqrt{\frac{x}{y}+\frac{y}{x}+2}+\frac{x}{y}+\frac{y}{x}>=\sqrt{2+2}+2=4\)
vp cua (1): \(2\left(x^2+y^2\right)>=\left(x+y\right)^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2>=2^2=4\)
dau = xay ra <=>x=y
tự giải tiếp