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a, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3.3-5=4\\x=\frac{33}{11}=3\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 3;4 )
b, Làm tương tự a
c, Ta có : \(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{15}{x-y+2}+\frac{10}{x+y-1}=20\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{29}{x-y+2}=29\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-y+2=1\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{3}{y-1-y+2}+\frac{2}{y-1+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\3+\frac{2}{2y-2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{2}{2y-2}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\2y-2=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-1=1\\y=2\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 1;2 )
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........
a) Xem lại đề
b) \(\left\{{}\begin{matrix}5x-3y=5\\2x+5y=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5.\frac{33-5y}{2}-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}165-25y-6y=10\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31y=155\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=\frac{33-5.5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=4\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\frac{x}{2}-\frac{y}{3}=0\\5x+y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{x}{2}-\frac{13-5x}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{3x-26+10x}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5.2\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
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:...
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)
\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
Đặt ẩn phụ: A=\(\frac{x}{4-3x}\);B=\(\frac{y}{2-y}\)
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}5A-3B=1\\3A-5B=-9\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}15A-9B=3\\15A-25B=-45\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}16B=48\\5A-3B=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}B=3\\5A-3.3=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}A=2\\B=3\end{matrix}\right.\)
Vậy hệ phương trình có một nghiệm duy nhất(x;y)=(2;3)
Ta có: \(\left\{{}\begin{matrix}\frac{x-y}{2}+\frac{x-3y}{4}=0\\\frac{3x-5y+1}{2}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x-y\right)+x-3y=0\\3x-5y+1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-5y=0\\3x-5y=1\end{matrix}\right.\)