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\(\hept{\begin{cases}2.\frac{1}{x}+5.\frac{1}{x+y}=2\\3.\frac{1}{x}+\frac{1}{x+y}=1,7\end{cases}}\)
Đặt \(\frac{1}{x}\)=a
\(\frac{1}{x+y}=b\)
ta có \(\hept{\begin{cases}2a+5b=2\\3a+b=1,7\end{cases}}\)
\(\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{5}\end{cases}}\)
=> \(\frac{1}{x}=\frac{1}{2}\Rightarrow x=2\)
\(\frac{1}{x+y}=\frac{1}{5}\)\(\Rightarrow x+y=5\)\(\Rightarrow y=3\)
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
Ta có hệ \(\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}=\frac{5}{2}\left(2\right)\end{cases}}\)
ĐK: \(x\ne0,y\ne0\)
Từ phương trình (2) ta có \(\frac{x^2y^2+1}{xy}=\frac{5}{2}\Rightarrow2x^2y^2-5xy+2=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{y}\\x=\frac{1}{2y}\end{cases}}\)
TH1: \(x=\frac{2}{y},\) thế vào phương trình (1) ta có:
\(\frac{2}{y}+y+\frac{y}{2}+\frac{1}{y}=\frac{9}{2}\Rightarrow\frac{3y}{2}+\frac{3}{y}=\frac{9}{2}\Rightarrow\frac{y}{2}+\frac{1}{y}=\frac{3}{2}\)
\(\Rightarrow\frac{y^2+2}{2y}=\frac{3}{2}\Rightarrow2y^2-6y+4=0\Rightarrow\orbr{\begin{cases}y=2\\y=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
TH2: \(x=\frac{1}{2y},\)
Thế vào phương trình (1) ta có:
\(\frac{1}{2y}+y+2y+\frac{1}{y}=\frac{9}{2}\Rightarrow3y+\frac{3}{2y}=\frac{9}{2}\Rightarrow y+\frac{1}{2y}=\frac{3}{2}\)
\(\Rightarrow\frac{2y^2+1}{2y}=\frac{3}{2}\Rightarrow4y^2-6y+2=0\Rightarrow\orbr{\begin{cases}y=1\\y=\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\) (Vô nghiệm)
Tóm lại, ta có 4 cặp nghiệm \(\left(1;2\right),\left(2;1\right),\left(1;\frac{1}{2}\right),\left(\frac{1}{2};1\right)\)
+) đặt \(a=x+\frac{1}{y};b=y+\frac{1}{x}\)
=> \(ab=\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=xy+\frac{1}{xy}+2=>xy+\frac{1}{xy}=ab-2\)
+) khi đó thay zô hệ phương trình ta đc
\(\hept{\begin{cases}a+b=\frac{9}{2}\\\frac{1}{4}+\frac{3}{2}a=ab-2\end{cases}\Rightarrow\hept{\begin{cases}2a+2b=9\\-4ab+6a+9=0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}2b=9-2a\\-2a\left(9-2a\right)+6a+9=0\end{cases}\Leftrightarrow\hept{\begin{cases}2b=9-2a\\4a^2-12a+9=0\end{cases}\Leftrightarrow}\hept{\begin{cases}2b=9-2a\\\left(2a-3\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=\frac{3}{2}\\b=3\end{cases}}}\)
+) trả zề biến x,y ta đc
\(\hept{\begin{cases}x+\frac{1}{y}=\frac{3}{2}\\y+\frac{1}{x}=3\end{cases}\Leftrightarrow\hept{\begin{cases}xy-\frac{3}{2}y+1=0\\xy-3x+1=0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(xy-\frac{3}{2}y+1\right)-\left(xy-3x+1\right)=0\\xy-3x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}y+3x=0\\xy-3x+1=0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}y=2x\\2x^2-3x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=2x\\2x^2-2x-x+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=2x\\\left(x-1\right)\left(2x-1=0\right)\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}hoặc\hept{\begin{cases}x=\frac{1}{2}\\y=1\end{cases}}}\)
+) thử lại ta thấy bộ số
\(\left(1;2\right);\left(\frac{1}{2};1\right)\)thỏa mãn hệ phương trình
zậy hệ phương trình có tập nghiệm (x,y) thuộc (1,2) ;(1/2 ;1)
Dat \(x+y=t;xy=v\left(t,v\ne0\right)\)
HPT tro thanh
\(\hept{\begin{cases}t+\frac{t}{v}=\frac{9}{2}\\v+\frac{1}{v}=\frac{5}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}t+\frac{t}{v}=\frac{9}{2}\left(1\right)\\v^2-\frac{5}{2}v+1=0\left(2\right)\end{cases}}\)
Xet (2):
\(\Delta=\frac{25}{4}-4=\frac{9}{4}\)
Suy ra:
\(v_1=4;v_2=1\)
Voi \(v=4\)thi thay vao HPT thay khong thoa man nen loai
Voi \(v=1\)thay vao HPT thay khong thoa man nen loai
Vay HPT vo nghiem
\(\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(\frac{1}{y}+y\right)=\frac{9}{2}\\\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=5\end{cases}}\)
dat an phu r giai