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28 tháng 6 2020
Điều kiện: \(y\ne0\); \(x+\frac{1}{y}\)\(\ge0\); \(x+y\ge3\).Đặt: \(a=\sqrt{x+\frac{1}{y}}\) ; \(b=\sqrt{x+y-3}\) \((a,b\ge0)\)Ta có: \(2x+y+\frac{1}{y}=8\)\(\ \Leftrightarrow\) \(\left(x+\frac{1}{y}\right)+\left(x+y-3\right)=5\) \(\Leftrightarrow\) \(a^2+b^2=5\)\(^{ }\)Hệ phương trình \(\Leftrightarrow\) \(\begin{cases}a+b=3\\a^2+b^2=5\end{cases}\)\(\ \Leftrightarrow\)\(\begin{cases}a=3-b\\\left(3-b\right)^2+b^2=5\ \left(1\right)\end{cases}\) \(\left(1\right)\Leftrightarrow\) \(b^2-6b+9+b^2=5\) \(\Leftrightarrow\) \(2b^2-6b+4=0\) \(\Leftrightarrow\) \(\Delta'=\left(-3\right)^2-2.4=1>0\) \(\Rightarrow\ \left(1\right)\) có hai nghiệm phân biệt: \(b_1=\frac{3+\sqrt{1}}{2}=2\) ; \(b_2=\frac{3-\sqrt{1}}{2}=1\) *Với \(b=2\) \(\Rightarrow\ a=3-b=3-2=1\)Ta có: \(\begin{cases}\sqrt{x+y-3}=2\\\sqrt{x+\frac{1}{y}}=1\ \end{cases}\)\(\ \) \(\Leftrightarrow\) \(\begin{cases}x+y-3=4\\x+\frac{1}{y}=1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}x+y=7\\x+\frac{1}{y}=1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}y-\frac{1}{y}=6\\x+y=7\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}y^2-6y-1=0\\x+y=7\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}\left[\begin{matrix}y=3+\sqrt{10}\\y=3-\sqrt{10}\end{matrix}\right.\\x=7-y\end{cases}\) \(\Leftrightarrow\) \(\left[\begin{matrix}\begin{cases}y=3+\sqrt{10}\\x=4-\sqrt{10}\end{cases}\\\begin{cases}y=3-\sqrt{10}\\x=4+\sqrt{10}\end{cases}\end{matrix}\right.\) *Với \(b=1\ \Rightarrow\ a=3-b=3-1=2\)Ta có: \(\begin{cases}\sqrt{x+y-3}=1\\\sqrt{x+\frac{1}{y}}=2\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}x+y=4\\x+\frac{1}{y}=4\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}y-\frac{1}{y}=0\\x+y=4\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}y^2-1=0\\x+y=4\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}\left[\begin{matrix}y=1\\y=-1\end{matrix}\right.\\x=4-y\end{cases}\) \(\Leftrightarrow\) \(\left[\begin{matrix}\begin{cases}y=1\\x=3\end{cases}\\\begin{cases}y=-1\\x=5\end{cases}\end{matrix}\right.\)
NV
18 tháng 2 2020

a/ \(\left\{{}\begin{matrix}\left(x^2+x\right)+\left(y^2+y\right)=18\\\left(x^2+x\right)\left(y^2+y\right)=72\end{matrix}\right.\)

Theo Viet đảo, \(x^2+x\)\(y^2+y\) là nghiệm của:

\(t^2-18t+72=0\Rightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=12\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\left\{2;-3\right\}\\y=\left\{3;-4\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{3;-4\right\}\\y=\left\{2;-3\right\}\end{matrix}\right.\end{matrix}\right.\)

NV
18 tháng 2 2020

b/ ĐKXĐ: ...

\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\x=\frac{3y-1}{y}\end{matrix}\right.\)

Nhận thấy \(y=\frac{1}{3}\) không phải nghiệm

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\\frac{1}{x}=\frac{y}{3y-1}\end{matrix}\right.\) \(\Rightarrow\frac{y}{3y-1}+\frac{1}{y+1}=1\)

\(\Leftrightarrow y\left(y+1\right)+3y-1=\left(3y-1\right)\left(y+1\right)\)

\(\Leftrightarrow y^2-y=0\Rightarrow\left[{}\begin{matrix}y=0\left(l\right)\\y=1\end{matrix}\right.\) \(\Rightarrow x=2\)

9 tháng 2 2020

a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)

Vậy..............................................................................

b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)

Vậy...................................................................................

c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)

\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)

Vậy hệ pt vô nghiệm

d) Nhân 3 pt đầu rồi thu gọn

20 tháng 6 2019

\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)

\(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

20 tháng 6 2019

\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)

Làm nốt nha

NV
10 tháng 7 2019

a/ Bạn tự giải

b/ ĐKXĐ:...

Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)

Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)

\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)

\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)

Chắc bạn ghi sai đề, nghiệm quá xấu

3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)

4/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)

\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)