Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có:
[tex]\left\{\begin{matrix} xy^{2}+x+y+\frac{1}{y}=4 & \\ y^{2}+x+\frac{1}{y}=3 & \end{matrix}\right.(y\neq 0)[/tex]
Từ phương trình suy ra:
[tex]\left\{\begin{matrix} y(xy+1)+\frac{xy+1}{y}=4 & \\ y^{2}+\frac{xy+1}{y}=3 & \end{matrix}\right.[/tex]
Đặt [tex]xy+1=a,y=b(b\neq 0)[/tex] ta có:
[tex]\left\{\begin{matrix} b^{2}+\frac{a}{b}=3 & \\ ab+\frac{a}{b}=4 & \end{matrix}\right.[/tex]
[tex]\Rightarrow \left\{\begin{matrix} 3b-b^{3}=a & \\ ab^{2}+a=4b & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3b-b^{3}=a & \\ b\left ( 2b^{2}-b^{4}-1 \right )=0 & \end{matrix}\right.[/tex]
[tex]\Leftrightarrow \left\{\begin{matrix} b=0 & \\ a=0 & \end{matrix}\right.[/tex](Loại) hoặc [tex]\left\{\begin{matrix} b=1 & \\ a=2 & \end{matrix}\right.[/tex] hoặc [tex]\left\{\begin{matrix} b=-1 & \\ a=-2 & \end{matrix}\right.[/tex]
TH1: [tex]\left\{\begin{matrix} b=1 & \\ a=2 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1 & \\ y=1 & \end{matrix}\right.[/tex]
TH2: [tex]\left\{\begin{matrix} b=-1 & \\ a=-2 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3 & \\ y=-1 & \end{matrix}\right.[/tex]
Vậy hệ phương trình có hai nghiệm: [tex]\left\{\begin{matrix} x=1 & \\ y=1 & \end{matrix}\right.[/tex] hoặc [tex]\left\{\begin{matrix} x=3 & \\ y=-1 & \end{matrix}\right.[/tex]
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình đề câu a phải như vậy nè:
\(a,\hept{\begin{cases}\frac{1}{x-2}+\frac{1}{y-1}=1\\\frac{2}{x-2}-\frac{3}{y-1}=1\end{cases}}\)\(Đkxđ:\hept{\begin{cases}x\ne2\\y\ne1\end{cases}}\)
Đặt: \(X=\frac{1}{x-2};Y=\frac{1}{y-1}\)
Ta có hệ sau:
\(\hept{\begin{cases}X+Y=1\\2X-3Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2\left(1-Y\right)-3Y=1\end{cases}}}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2-5Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=\frac{4}{5}\\Y=\frac{1}{5}\end{cases}}}\)
Với \(X=\frac{4}{5}\Rightarrow\frac{1}{x-2}=\frac{4}{5}\Leftrightarrow4\left(x-2\right)=5\Leftrightarrow x=\frac{13}{4}\)
Với \(Y=\frac{1}{5}\Rightarrow\frac{1}{y-1}=\frac{1}{5}\Leftrightarrow y-1=5\Leftrightarrow y=6\)
Vậy nghiệm của hệ pt là: \(\left(x;y\right)=\left(\frac{13}{4};6\right)\)
Câu b e nghĩ đề như vậy nè:
\(b,\hept{\begin{cases}\frac{7}{\sqrt{x-7}}-\frac{4}{\sqrt{y+6}}=\frac{5}{3}\\\frac{5}{\sqrt{x-7}}+\frac{3}{\sqrt{y+6}}=\frac{3}{6}\end{cases}}\) \(Đkxđ:\hept{\begin{cases}x>7\\x>-6\end{cases}}\)
Đặt \(\frac{1}{\sqrt{x-7}}=a\left(a>0\right);\frac{1}{\sqrt{y+6}}=b\left(b>0\right)\)
Ta có hệ pt mới: \(\hept{\begin{cases}7a-4b=\frac{5}{3}\\5a+3b=\frac{13}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\end{cases}}\left(tmđk\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-7}}=\frac{1}{3}\\\frac{1}{\sqrt{y+6}}=\frac{1}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x-7=9\\x+6=36\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16\\y=30\end{cases}\left(tmđk\right)}\)
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(16;30\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\hept{\begin{cases}\frac{x}{\sqrt{y}}+\frac{2\sqrt{y}}{x}=\frac{2}{x}+\frac{1}{\sqrt{y}}-3\left(1\right)\\x^2-xy-9x+12=0\left(2\right)\end{cases}}\)
Đặt \(\frac{2}{x}=a,\frac{1}{\sqrt{y}}=b\left(b>0\right)\)
\(\left(1\right)\Leftrightarrow\frac{2b}{a}+\frac{a}{b}=a+b-3\)
\(\Leftrightarrow2b^2+a^2+3ab=ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a+2b\right)=\left(a+b\right)ab\)
\(\Leftrightarrow\left(a+b\right)\left(a-ab+2b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-b\left(3\right)\\a-ab+2b=0\left(4\right)\end{cases}}\)
Giải (3)
\(\left(3\right)\Leftrightarrow\frac{2}{x}=-\frac{1}{\sqrt{y}}\Leftrightarrow\frac{4}{x^2}=\frac{1}{y}\)
\(\Leftrightarrow y=\frac{x^2}{4}\). Thay vào (2) tìm nghiệm (x,y)
Giải (4)
\(\left(4\right)\Leftrightarrow\frac{2}{x}-\frac{2}{\sqrt{y}}+\frac{2}{x\sqrt{y}}=0\)
\(\Leftrightarrow\sqrt{y}-x+2=0\)
Giải tiếp là ra
Học tốt!!!!!!!!!