Bài 1 : giải các phương trình sau
1 , \(\left(x^2-6x\right)\sqrt{17-x^2}=x^2-6x\)
2 , \(\left(x^2+5x+4\right)\sqrt{x+3}=0\)
3, \(\sqrt{3x}+\sqrt{2x-2}=\sqrt{1-x}+2\)
4, \(\left(x^2-4x+3\right)\sqrt{x-2}=0\)
5 , \(\sqrt{x^2+3x-2}=\sqrt{1+x}\)
6 , \(\left(\sqrt{x-4}-1\right)\left(x^2-7x+6\right)=0\)
7, \(\sqrt{2x^2-8x+4}=x-2\)
8 , \(\sqrt{3x+7}-\sqrt{x+1}=2\)
a/ ĐKXĐ: ...
\(\Leftrightarrow\left(x^2-6x\right)\left(\sqrt{17-x^2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x=0\\\sqrt{17-x^2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-6\right)=0\\x^2=16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\left(l\right)\\x=4\\x=-4\end{matrix}\right.\)
b/ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=0\\\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\left(l\right)\\x=-3\end{matrix}\right.\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ge1\\x\le1\end{matrix}\right.\) \(\Rightarrow x=1\)
Thay \(x=1\) vào pt thấy ko thỏa mãn
Vậy pt vô nghiệm
d/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\\sqrt{x-2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\left(l\right)\\x=2\end{matrix}\right.\)
e/ ĐKXĐ: \(x\ge\frac{-3+\sqrt{17}}{2}\)
\(\Leftrightarrow x^2+3x-2=x+1\)
\(\Leftrightarrow x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\left(l\right)\end{matrix}\right.\)
f/ ĐKXĐ: \(x\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-4}-1=0\\x^2-7x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=1\\\left(x-1\right)\left(x-6\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\left(l\right)\\x=6\end{matrix}\right.\)
g/ \(x\ge2\)
\(\Leftrightarrow2x^2-8x+4=\left(x-2\right)^2\)
\(\Leftrightarrow2x^2-8x+4=x^2-4x+4\)
\(\Leftrightarrow x^2-4x=0\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=4\end{matrix}\right.\)
h/ ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\sqrt{3x+7}=2+\sqrt{x+1}\)
\(\Leftrightarrow3x+7=x+5+4\sqrt{x+1}\)
\(\Leftrightarrow x+1-2\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
