1.
a, Tham khảo:
Câu hỏi của Rell - Toán lớp 10 | Học trực tuyến
b, \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
ĐKXĐ: \(x\ge1;x\le-3;x=-1\)
Ta có \(2x+2=\sqrt{2x^2+8x+6}+\sqrt{x^2-1}\ge0\Rightarrow x+1\ge0\)
\(pt\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
\(\Leftrightarrow\sqrt{x+1}.\left(\sqrt{2x+6}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{2x+6}+\sqrt{x-1}=2\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\\sqrt{2x+6}+\sqrt{x-1}=2\sqrt{x+1}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+6+x-1+2\sqrt{\left(2x+6\right)\left(x-1\right)}=4\left(x+1\right)\)
\(\Leftrightarrow2\sqrt{\left(2x+6\right)\left(x-1\right)}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(2x+6\right)\left(x-1\right)=\left(x-1\right)^2\\x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(-7x-25\right)=0\\x\ge1\end{matrix}\right.\Leftrightarrow x=1\left(tm\right)\)
Vậy ...
c, \(\sqrt{4x-9}-\sqrt{11x+1}-\sqrt{7x+4}=0\)
ĐKXĐ: \(x\ge\frac{9}{4}\)
\(pt\Leftrightarrow\sqrt{4x-9}=\sqrt{11x+1}+\sqrt{7x+4}\)
\(\Leftrightarrow4x-9=11x+1+7x+4+2\sqrt{\left(11x+1\right)\left(7x+4\right)}\)
\(\Leftrightarrow2\sqrt{\left(11x+1\right)\left(7x+4\right)}=-14x-14\left(2\right)\)
Do \(x\ge\frac{9}{4}\Rightarrow-14x-14< 0\Rightarrow\left(2\right)\) vô nghiệm