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gọi \(\frac{1}{2x-y}\)là \(a\); \(\frac{1}{x-2y}\)là \(b\)
Ta có hệ phương trình: \(\hept{\begin{cases}2a+3b=\frac{1}{2}\\2a-b=\frac{1}{18}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=\frac{1}{12}\\b=\frac{1}{9}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{2x-y}=\frac{1}{12}\\\frac{1}{x-2y}=\frac{1}{9}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x-y=12\\x-2y=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)
Đặt x +\(\frac{1}{x}\) =a, y+\(\frac{1}{y}\)=b
| hpt<=>\(\hept{\begin{cases}a^2-2+b^2-2=1\\a+b=3\end{cases}}\) | |
|---|---|
| đến đây thì dễ rồi , có tổng với tích | |
| bạn tìm ra a,b rồi tương tự tìm x,y |
\(1,\hept{\begin{cases}x\left(x+y+1\right)=3\\\left(x+y\right)^2-\frac{5}{x^2}=-1\end{cases}\left(ĐKXĐ:x\ne0\right)}\)
\(\Leftrightarrow\hept{\begin{cases}x+y=\frac{3}{x}-1\\\left(x+y\right)^2-\frac{5}{x^2}=-1\end{cases}}\)
\(\Rightarrow\left(\frac{3}{x}-1\right)^2-\frac{5}{x^2}=-1\)
Đặt \(\frac{1}{x}=a\left(a\ne0\right)\)
\(\Rightarrow\left(3a-1\right)^2-5a^2=-1\)
\(\Leftrightarrow9a^2-6a+1-5a^2+1=0\)
\(\Leftrightarrow4a^2-6a+2=0\)
Làm nốt
2, ĐKXĐ \(x\ge1,y\ge0\)
\(\hept{\begin{cases}xy+x+y=x^2-2y^2\left(1\right)\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\left(2\right)\end{cases}}\)
Pt (1) <=> \(xy+x+y+y^2=x^2-y^2\)
<=> \(y\left(x+y\right)+x+y=\left(x-y\right)\left(x+y\right)\)
<=> \(\left(x+y\right)\left(y+1\right)=\left(x-y\right)\left(x+y\right)\)
<=> \(\left(x+y\right)\left(2y+1-x\right)=0\)
Mà \(x\ge1,y\ge0\) => \(x+y>0\) => \(2y+1-x=0\)<=> \(x=2y+1\)
Thay x=2y+1 vào (2)
Đoạn này bn tự giải tiếp nhé
2 \(\hept{\begin{cases}\frac{x^2+1}{y}=\frac{y^2+1}{y}\left(1\right)\\x^2+3y^2=4\left(2\right)\end{cases}}\)
ĐK \(x,y\ne0\)
Từ \(\frac{y^2+1}{y}=\frac{x^2+1}{x}\Leftrightarrow xy^2+x=x^2y+y\Leftrightarrow\left(xy-1\right)\left(x-y\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=y\\xy=1\end{cases}}\)
+ thay \(x=y\)vào (2) ta dc ..................
+xy=1 suy ra 1=1/y thay vao 2 ta dc............
ta có điều kiện \(x\ne0;y\ne0\)ta có
\(\frac{1}{x^3}+\frac{1}{y^3}+3=x^3y^3\Leftrightarrow\left(\frac{1}{x}\right)^3+\left(\frac{1}{y}\right)^3+\left(-x^3y^3\right)=3.\frac{1}{x}.\frac{1}{y}.\left(-xy\right)\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{1}{y}=-xy\\\frac{1}{x}+\frac{1}{y}-xy=0\end{cases}}\)
TH1 : ta có \(\frac{1}{x}=\frac{1}{y}=-xy\Leftrightarrow\hept{\begin{cases}x=y\\1=-x^2y\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}}\)(thử zô (1) ko thỏa mãn )
TH2 :ta có \(\frac{1}{x}+\frac{1}{y}-xy=0\Leftrightarrow x+y=\left(xy\right)^2\)ta có
\(\left(1\right)\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}=3+x^2y^2\Leftrightarrow xy\left(3xy+2\right)=0\Leftrightarrow xy=-\frac{2}{3}\)
\(\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}=3+x^2y^2\left(1\right)\\\frac{1}{x^3}+\frac{1}{y^3}+3=x^3y^3\left(2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}=3+x^2y^2\\\orbr{\begin{cases}\frac{1}{x}=\frac{1}{y}=-xy\\\frac{1}{x}+\frac{1}{y}-xy=0\end{cases}}\end{cases}}}\)zậy \(\hept{\begin{cases}x+y=\left(xy\right)^2\\xy=-\frac{2}{3}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2+\sqrt{58}}{9}\\y=\frac{2-\sqrt{58}}{9}\end{cases}hoặc\hept{\begin{cases}x=\frac{2-\sqrt{58}}{9}\\y=\frac{2+\sqrt{58}}{9}\end{cases}}}}\)
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
\(\Leftrightarrow\hept{\begin{cases}x^3y^3+1=2y^2\left(1\right)\\\frac{x^2}{y^2}+\frac{x}{y^2}=2\left(2\right)\end{cases}}\)
(1) : (xy + 1) (x2y2 + xy +1 )=2y2
(2) : \(\frac{x}{y}\)( x + \(\frac{1}{y}\)) = 2
\(\frac{x\left(xy+1\right)}{y^2}\)=2
x(xy +1 ) =2y2
\(\Rightarrow\)x(xy +1 )=( x2y2 + xy +1 ). (xy +1 )
(xy + 1 ) (x - x2y2 - xy - 1 ) = 0
\(\orbr{\begin{cases}xy+1=0\\x-x^2y^2-xy-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}xy=-1\\x-x^2y^2-xy-1=0\end{cases}}\)
tự giải tiếp nha
#mã mã#
ĐKXĐ : \(x\ne-2;y\ne\frac{3}{2}\)
Đặt : \(a=\frac{1}{x+2};b=\frac{1}{2y-3}\)
HPT đã cho : \(\hept{\begin{cases}2a-b=2\\6a-2b=1\end{cases}\Leftrightarrow\hept{\begin{cases}a=-\frac{3}{2}\\b=-5\end{cases}}}\) hay \(\hept{\begin{cases}\frac{1}{x+2}=-\frac{3}{2}\\\frac{1}{2y-3}=-5\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{8}{3}\\y=\frac{7}{5}\end{cases}}}\)
Vậy ...
ĐK : x khác -2 ; y khác 3/2
Đặt \(\hept{\begin{cases}a=\frac{2}{x+2}\\b=\frac{1}{2y-3}\end{cases}\left(a,b\ne0\right)}\)
hpt đã cho trở thành \(\hept{\begin{cases}a-b=2\\3a-2b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2+b\\b=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-3\\b=-5\end{cases}\left(tm\right)}\)
\(\Rightarrow\hept{\begin{cases}\frac{2}{x+2}=-3\\\frac{1}{2y-3}=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{8}{3}\\y=\frac{7}{5}\end{cases}\left(tm\right)}\)