Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có hệ pt :
\(\left\{{}\begin{matrix}\frac{y}{3}=\frac{x}{4}\\\frac{z}{9}=\frac{x}{4}\\7x-3y+2z=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\7x-\frac{3.3x}{4}+\frac{2.9x}{4}=90\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\28x-9x+18x=360\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\37x=360\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{4}.\frac{360}{37}\\z=\frac{9}{4}.\frac{360}{37}\\x=\frac{360}{37}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{270}{37}\\z=\frac{810}{37}\\x=\frac{360}{37}\end{matrix}\right.\)
Vậy . . . . . . . . .
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
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, \(x^2+x\) và \(y^2+y\) là nghiệm của:
\(t^2-18t+72=0\Rightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=12\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\left\{2;-3\right\}\\y=\left\{3;-4\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{3;-4\right\}\\y=\left\{2;-3\right\}\end{matrix}\right.\end{matrix}\right.\)
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
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\\frac{1}{x}=\frac{y}{3y-1}\end{matrix}\right.\) \(\Rightarrow\frac{y}{3y-1}+\frac{1}{y+1}=1\)
\(\Leftrightarrow y\left(y+1\right)+3y-1=\left(3y-1\right)\left(y+1\right)\)
\(\Leftrightarrow y^2-y=0\Rightarrow\left[{}\begin{matrix}y=0\left(l\right)\\y=1\end{matrix}\right.\) \(\Rightarrow x=2\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)
\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
Bài 1:
Lấy PT $(1)$ trừ PT $(2)$ ta có:
\(x^2-y^2=3y-3x\)
\(\Leftrightarrow (x-y)(x+y)+3(x-y)=0\Leftrightarrow (x-y)(x+y+3)=0\)
$\Rightarrow x-y=0$ hoặc $x+y+3=0$
Nếu $x-y=0\Leftrightarrow x=y$. Thay vào PT $(1)$:
\(x^2=3x-2\Leftrightarrow x^2-3x+2=0\Leftrightarrow (x-1)(x-2)=0\)
$\Rightarrow x=1$ hoặc $x=2$
Tương ứng ta thu được $y=1$ hoặc $y=2$
Nếu $x+y+3=0\Leftrightarrow y=-(x+3)$. Thay vào PT $(1)$:
\(x^2=-3(x+3)-2\Leftrightarrow x^2=-3x-11\Leftrightarrow x^2+3x+11=0\)
\(\Leftrightarrow (x+\frac{3}{2})^2=\frac{-35}{4}< 0\) (vô lý)
Vậy..........
Bài 2:
Lấy PT(1) trừ PT(2) ta có:
\(2x-2y+\frac{1}{y}-\frac{1}{x}=\frac{3}{x}-\frac{3}{y}\)
\(\Leftrightarrow 2(x-y)+(\frac{4}{y}-\frac{4}{x})=0\)
\(\Leftrightarrow (x-y)+\frac{2(x-y)}{xy}=0\)
\(\Leftrightarrow (x-y).\frac{2+xy}{xy}=0\Rightarrow \left[\begin{matrix} x=y\\ xy=-2\end{matrix}\right.\)
Nếu $x=y$. Thay vào PT (1) có:
\(2x+\frac{1}{x}=\frac{3}{x}\Leftrightarrow 2x-\frac{2}{x}=0\Leftrightarrow x^2-1=0\)
\(\Rightarrow x^2=1\Rightarrow x=\pm 1\Rightarrow y=\pm 1\) (tương ứng)
Nếu $xy=-2\Rightarrow \frac{1}{y}=\frac{-x}{2}$
Thay vào PT(1): $2x-\frac{x}{2}=\frac{3}{x}$
$\Leftrightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$
$\Rightarrow y=\mp \sqrt{2}$
Vậy........
\(\left\{{}\begin{matrix}7x-3y=5\\\frac{x}{2}+\frac{y}{3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+5}{7}\\\frac{3y+5}{14}+\frac{y}{3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy...