bạn thiếu đề rồi, phải cần thêm ptđt nữa để tìm trục tung mới có thể tìm m được nhé

Giải được mà bạn 

Do (d) :y=(m-2)x+m cắt trục hoành tại điểm x=-2 nên đường thẳng đi qua điểm (-2:0)

\(\Rightarrow0=\left(m-2\right).\left(-2\right)+m\)

\(\Leftrightarrow0=-2m+4+m\)

\(\Leftrightarrow m=4\)

Vậy m=4 thì thỏa mãn 

1. \(\sqrt{x^2-4}-x^2+4=0\)( ĐK: \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\))

\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)

\(\Leftrightarrow\left(x^2-4\right)^2=x^2-4\)

\(\Leftrightarrow\left(x^2-4\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\left(tm\right)\\x=\pm\sqrt{5}\left(tm\right)\end{cases}}\)

Vậy pt có tập no \(S=\left\{2;-2;\sqrt{5};-\sqrt{5}\right\}\)

2. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)ĐK: \(\hept{\begin{cases}x^2-4x+5\ge0\\x^2-4x+8\ge0\\x^2-4x+9\ge0\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2-4x+5}-1+\sqrt{x^2-4x+8}-2+\sqrt{x^2-4x+9}-\sqrt{5}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{\sqrt{x^2-4x+5}+1}+\frac{x^2-4x+4}{\sqrt{x^2-4x+8}+2}+\frac{x^2-4x+4}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}\right)=0\)

Từ Đk đề bài \(\Rightarrow\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}>0\)

\(\Rightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy pt có no x=2

mình muốn âm điểm

\(a,x-\sqrt{x-4\sqrt{x}+4}=8\)

\(x-\sqrt{\left(\sqrt{x}-2\right)^2}=8\)

\(x-\left|\sqrt{x}-2\right|=8\)

\(TH1:0\le x\le2\)

\(x-2+\sqrt{x}=8\)

\(x+\sqrt{x}-10=0\)

\(\sqrt{\Delta}=1-\left(4.-10\right)=\sqrt{41}\)

\(\orbr{\begin{cases}x_1=\frac{\sqrt{41}-1}{2}\left(KTM\right)\\x_2=\frac{-\sqrt{41}-1}{2}\left(KTM\right)\end{cases}}\)

\(TH2:x>2\)

\(x-\sqrt{x}+2=8\)

\(x-\sqrt{x}-6=0\)

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

\(\orbr{\begin{cases}\sqrt{x}+2=0\\\sqrt{x}-3=0\end{cases}\orbr{\begin{cases}\sqrt{x}=-2\left(KTM\right)\\x=9\left(TM\right)\end{cases}}}\)

\(b,\sqrt{\frac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}}=0\)

\(\sqrt{\left(\frac{1}{2}x+1\right)^2}-\sqrt{\sqrt{5}^2-2\sqrt{5}+1}=0\)

\(\left|\frac{1}{2}x+1\right|-\sqrt{\left(\sqrt{5}-1\right)^2}=0\)

\(\left|\frac{1}{2}x+1\right|-\sqrt{5}+1=0\)

\(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)

\(\orbr{\begin{cases}\frac{1}{2}x+1=\sqrt{5}-1\\\frac{1}{2}x+1=1-\sqrt{5}\end{cases}\orbr{\begin{cases}\frac{1}{2}x=\sqrt{5}-2\\\frac{1}{2}x=-\sqrt{5}\end{cases}\orbr{\begin{cases}x=2\sqrt{5}-4\\x=-2\sqrt{5}\end{cases}}}}\)

\(c,\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)

\(\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)

\(\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}+1\right|=4\)

\(\sqrt{2x-5}+3+\sqrt{2x-5}+1=4\)

\(\sqrt{2x-5}=0\)

\(x=\frac{5}{2}\left(TM\right)\)

\(x-\sqrt{x-2\sqrt{x}2+2^2}=8.\)

\(x-\sqrt{\left(\sqrt{x}-2\right)^2}=8\)

\(x-\sqrt{x}+2=8\)

\(\sqrt{x}\left(\sqrt{x}-1\right)=6\)

\(\Leftrightarrow\sqrt{x}=3,\sqrt{x}-1=2\Leftrightarrow x=3^2=9\)

\(\frac{\sqrt{35}+\sqrt{14}}{\sqrt{2}+\sqrt{5}}-\frac{2\sqrt{21}-\sqrt{56}}{\sqrt{3}-\sqrt{2}}\)

\(\frac{\sqrt{7}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}}-\frac{2\sqrt{7}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}\)

\(=\sqrt{7}-2\sqrt{7}\)

\(=-\sqrt{7}\)