có ai giúp ko nè

Câu 1:

\(\left\{{}\begin{matrix}m^2x+y=3m\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2x-4x=3m+6\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(m^2-4\right)=3m+6\\-4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+6}{m^2-4}=\dfrac{3}{m-2}\\y=6-\dfrac{3}{m-2}=\dfrac{6m-15}{m-2}\end{matrix}\right.\)Câu 2:

\(\left\{{}\begin{matrix}5x-y=13\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15x-3y=39\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}16x=32\\x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

Anh vẫn thấy bình thường mà em, vẫn thấy cách làm

Bài 1: 

\(P=\left(\dfrac{x-\sqrt{x}-2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{-2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)

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

\(\Rightarrow x+y+x-y=m-1+m+3\)

\(\Rightarrow2x=2m+2\Rightarrow x=m+1\)

\(\Rightarrow x_0=m+1\) (1)

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

\(\Rightarrow x+y-\left(x-y\right)=m-1-\left(m+3\right)\)

\(\Rightarrow2y=-4\Rightarrow y=-2\Rightarrow y_0=-2\Rightarrow y_0^2=4\) (2)

-Từ (1) và (2) suy ra:

\(m+1=4\Rightarrow m=3\)

lớp 8 mà có rút gọn có căn r à :V

Đề bài yêu cầu gì?

\(Dk:x,y\ge\frac{-5}{4}\)

\(\left\{{}\begin{matrix}\left(2x-3\right)^2=4y+5\\\left(2y-3\right)^2=4x+5\end{matrix}\right.\Rightarrow\left(2y-3\right)^2-\left(2x-3\right)^2=4x-4y\Leftrightarrow\left(2y-2x\right)\left(2x+2y-6\right)=4\left(x-y\right)\Leftrightarrow4\left(y-x\right)\left(x+y-3\right)=4\left(x-y\right)\Leftrightarrow-4\left(x-y\right)\left(x+y-3\right)=4\left(x-y\right)\)

\(+,x=y\Rightarrow\left(2x-3\right)^2=4x+5\Leftrightarrow4x^2-12x+9=4x+5\Leftrightarrow4x^2-16x+4=0\Leftrightarrow x^2-4x+1=0\)

\(\Delta=16-4=12>0\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y=2+\sqrt{3}\left(tm\right)\\x=y=2-\sqrt{3}\left(tm\right)\end{matrix}\right.\)

\(+,x\ne y\Rightarrow-4\left(x+y-3\right)=4\Leftrightarrow x+y-3=-1\Leftrightarrow x+y=2\)

\(\Leftrightarrow x=2-y\Rightarrow\left(1-2y\right)^2=4y+5\Leftrightarrow1-4y+4y^2=4y+5\Leftrightarrow4y^2-8y-4=0\Leftrightarrow y^2-2y-1=0;\Delta=\left(-2\right)^2-\left(-1\right).1.4=4-\left(-4\right)=8>0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1-\sqrt{2};x=1+\sqrt{2}\left(tm\right)\\x=1-\sqrt{2};y=1+\sqrt{2}\left(tm\right)\end{matrix}\right.\)

Để hệ vô nghiệm thì \(\dfrac{m^2}{-4}=\dfrac{1}{-1}\ne\dfrac{3m}{6}=\dfrac{m}{2}\)

=>\(\left\{{}\begin{matrix}\dfrac{m^2}{-4}=\dfrac{1}{-1}=-1\\\dfrac{m}{2}\ne-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m^2=4\\m\ne-2\end{matrix}\right.\)

=>m=2

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

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4\\6x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-14\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{14}{11}\\x=\dfrac{y+2}{3}=\dfrac{\dfrac{14}{11}+2}{3}=\dfrac{12}{11}\end{matrix}\right.\)

\(9,\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\2x+3\left(3x-2\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\11x=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{12}{11}\\y=\dfrac{14}{11}\end{matrix}\right.\)

\(10,\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\2\left(2-3y\right)-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\4-6y-y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{14}\\y=\dfrac{3}{7}\end{matrix}\right.\)

1) \(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\8x+3\left(2-4x\right)=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=1\end{matrix}\right.\)

2) 2 pt 3 ẩn không giải được.

3) \(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\3x+2\left(x-2\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

4) \(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+1}{2}\\-4\cdot\frac{3y+1}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)

5) \(\left\{{}\begin{matrix}2x+3y=5\\5x-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-3y+5}{2}\\5\cdot\frac{-3y+5}{2}-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)

6) \(\left\{{}\begin{matrix}3x-y=7\\x+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2\left(3x-7\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

7) \(\left\{{}\begin{matrix}x+4y=2\\3x+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2-4y\\3\left(2-4y\right)+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{1}{5}\\x=\frac{6}{5}\end{matrix}\right.\)

8) \(\left\{{}\begin{matrix}-x-y=2\\-2x-3y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-x-2\\-2x-3\left(-x-2\right)=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)

9) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)