\(\left\{{}\begin{matrix}mx-y=2\\3x+my=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2x-my=2m\\3x+my=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+3\right)x=2m+5\\y=mx-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=mx-2\end{matrix}\right.\)

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

Thay vào \(x+y=1-\dfrac{m^2}{m^2+3}\)

\(\Leftrightarrow\dfrac{3m+5}{m^2+3}+\dfrac{5m-6}{m^2+3}=1-\dfrac{m^2}{m^2+3}\)

\(\Leftrightarrow\dfrac{8m-1}{m^2+3}=\dfrac{3}{m^2+3}\)

\(\Leftrightarrow8m-1=3\)

\(\Rightarrow m=\dfrac{1}{2}\)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge\dfrac{1}{\sqrt{2}}\\x\le-\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)

 Pt\(\Leftrightarrow8x^2-4-2\left(3x+1\right)\sqrt{2x^2-1}+2x^2+3x-2=0\)

\(\Leftrightarrow4\left(2x^2-1\right)-2\left(3x+1\right)\sqrt{2x^2-1}+2x^2+3x-2=0\)

Đặt \(\sqrt{2x^2-1}=t\)

\(\Rightarrow4t^2-2\left(3x+1\right)t+2x^2+3x-2=0\)

Coi pt trên là pt bậc 2 ẩn t tham số x, ta có:

\(\Delta'=\left(3x+1\right)^2-4\left(2x^2+3x-2\right)=x^2-6x+9=\left(x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1+x-3}{4}=\dfrac{2x-1}{2}\\t=\dfrac{3x+1-\left(x-3\right)}{4}=\dfrac{x+2}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2-1}=\dfrac{2x-1}{2}\\\sqrt{2x^2-1}=\dfrac{x+2}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{2x^2-1}=2x-1\left(\text{với }x\ge\dfrac{1}{2}\right)\\2\sqrt{2x^2-1}=x+2\left(\text{với }x\ge-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(2x^2-1\right)=\left(2x-1\right)^2\left(\text{với }x\ge\dfrac{1}{2}\right)\\4\left(2x^2-1\right)=\left(x+2\right)^2\left(\text{với }x\ge-2\right)\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}4x^2+4x-5=0\left(\text{với }x\ge\dfrac{1}{2}\right)\\7x^2-4x-8=0\left(\text{với }x\ge-2\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{6}}{2}\\x=\dfrac{-1-\sqrt{6}}{2}< \dfrac{1}{2}\left(loại\right)\\x=\dfrac{2+2\sqrt{15}}{7}\\x=\dfrac{2-2\sqrt{15}}{7}\end{matrix}\right.\)

Công thức: 

9.

\(\Delta=\left(m-2\right)^2+32>0;\forall m\Rightarrow\) pt luôn có 2 nghiệm pb với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-2\\x_1x_2=-8\end{matrix}\right.\)

\(Q=\left(x_1^2-1\right)\left(x_2^2-1\right)=\left(x_1x_2\right)^2-\left(x_1^2+x_2^2\right)+1\)

\(=\left(x_1x_2\right)^2-\left(x_1+x_2\right)^2+2x_1x_2+1\)

\(=\left(-8\right)^2-\left(m-2\right)^2+2.\left(-8\right)+1\)

\(=49-\left(m-2\right)^2\)

Do \(\left(m-2\right)^2\ge0;\forall m\)

\(\Rightarrow49-\left(m-2\right)^2\le49\)

Dấu "=" xảy ra khi \(m-2=0\Rightarrow m=2\)

10.

\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\)

\(\Rightarrow\) Pt luôn có 2 nghiệm với mọi m

Theo hệ thức VietL \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)

kết hợp điều kiện đề bài và hệ thức Viet ta được:

\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1-x_2=2\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x_1=m+2\sqrt{5}\\x_2=x_1-2\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{m+2\sqrt{5}}{2}\\x_2=\dfrac{m-2\sqrt{5}}{2}\end{matrix}\right.\)

Thế vào \(x_1x_2=m-2\)

\(\Rightarrow\left(\dfrac{m+2\sqrt{5}}{2}\right)\left(\dfrac{m-2\sqrt{5}}{2}\right)=m-2\)

\(\Leftrightarrow m^2-20=4m-8\)

\(\Leftrightarrow m^2-4m-12=0\Rightarrow\left[{}\begin{matrix}m=6\\m=-2\end{matrix}\right.\)

Hệ có nghiệm duy nhất khi: \(\dfrac{3}{1}\ne\dfrac{m}{-2}\Rightarrow m\ne-6\)

Khi đó ta có:

\(\left\{{}\begin{matrix}3x+my=5\\x-2y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x+2my=10\\mx-2my=3m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+6\right)x=3m+10\\y=\dfrac{x-3}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{x-3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{-4}{m+6}\end{matrix}\right.\)

\(2x+y=1\Rightarrow\dfrac{2\left(3m+10\right)}{m+6}+\dfrac{-4}{m+6}=1\)

\(\Leftrightarrow\dfrac{6m+16}{m+6}=1\)

\(\Rightarrow6m+16=m+6\)

\(\Rightarrow m=-2\)

ĐKXĐ: \(x>0\)

\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\dfrac{x^2+5x+6}{x}}\)

\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\dfrac{\left(x+2\right)\left(x+3\right)}{x}}\)

\(\Leftrightarrow\sqrt{x\left(x+3\right)}-\sqrt{\dfrac{\left(x+2\right)\left(x+3\right)}{x}}+2\sqrt{x+2}-2x=0\)

\(\Leftrightarrow\sqrt{\dfrac{x+3}{x}}\left(x-\sqrt{x+2}\right)-2\left(x-\sqrt{x+2}\right)=0\)

\(\Leftrightarrow\left(\sqrt{\dfrac{x+3}{x}}-2\right)\left(x-\sqrt{x+2}\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+3}{x}}=2\\x=\sqrt{x+2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+3}{x}=2\\x^2=x+2\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\\x=-1< 0\left(loại\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{4x^2-8x+4}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{\left(2x-2\right)^2}+5\sqrt{\left(y+2\right)^2}=13\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2.2\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\\ \left|x-1\right|=a;\left|y+2\right|=b\\ \Rightarrow\left\{{}\begin{matrix}5a-3b=7\\4a+5b=13\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}25a-15b=35\\12a+15b=39\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}37a=74\\5a-3b=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=2\\5.2-3b=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\\\left[{}\begin{matrix}y+2=1\\y+2=-1\end{matrix}\right.\end{matrix}\right.\\ \left\{{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}y=-1\\y=-3\end{matrix}\right.\end{matrix}\right.\)

bạn thêm dấu \("\Leftrightarrow"\) ở cuối nhé

7.

\(\Delta'=9-\left(m+4\right)=5-m\)

Pt có 2 nghiệm khi: \(5-m\ge0\Rightarrow m\le5\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=6\\x_1x_2=m+4\end{matrix}\right.\)

\(2020\left(x_1+x_2\right)-2021x_1x_2=2014\)

\(\Leftrightarrow2020.6-2021\left(m+4\right)=2014\)

\(\Leftrightarrow m=\dfrac{2022}{2021}\) (thỏa mãn)

8.

\(\Delta'=m^2+1>0;\forall m\Rightarrow pt\) luôn có 2 nghiệm pb với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=-1\end{matrix}\right.\)

\(x_1^2+x_2^2-x_1x_2=7\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-3x_1x_2=7\)

\(\Leftrightarrow4m^2+3=7\)

\(\Leftrightarrow m^2=1\)

\(\Rightarrow m=\pm1\)