\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)\(\left(đk:\dfrac{3}{2}\le x\le\dfrac{5}{2}\right)\)
\(VT\le\sqrt{2\left(2x-3+5-2x\right)}=\sqrt{2.2}=2\)\(\left(bunhiacopxki\right)\)
\(VP=3\left(x-2\right)^2+2\ge2\)
\(\Rightarrow dấu"="\Leftrightarrow\left\{{}\begin{matrix}x=2\\2x-3=5-2x\end{matrix}\right.\)\(\Leftrightarrow x=2\left(thỏa\right)\)
\(\left(đk:-\dfrac{1}{3}\le x\le6\right)\sqrt{6-x}-\sqrt{3x+1}=3x^2-14x-8\)
2.xem lại đề
\(3;đặt:\sqrt{2x+1}=y\Rightarrow y^2=2x+1\left(1\right)\)
\(x^2-1=2\sqrt{2x+1}\Rightarrow x^2-1=2y\Leftrightarrow x^2=2y+1\left(2\right)\)
\(lấy\left(2\right)-\left(1\right)\Rightarrow x^2-y^2=2y+1-2x-1\Leftrightarrow\left(x-y\right)\left(x+y+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\left(3\right)\\x=-\left(y+2\right)\left(4\right)\end{matrix}\right.\)
thay(3) và (4) lần lượt vào(1) ta giải ra được x
b) ĐKXĐ : \(-\dfrac{1}{3}\le x\le6\)
Ta có \(\sqrt{6-x}-\sqrt{3x+1}=3x^2-14x-8\)
<=> \(\left(\sqrt{6-x}-1\right)+\left(4-\sqrt{3x+1}\right)=3x^2-14x-5\)
<=> \(\dfrac{5-x}{\sqrt{6-x}+1}+\dfrac{15-3x}{4+\sqrt{3x+1}}=\left(x-5\right)\left(3x+1\right)\)
\(\Leftrightarrow\left(x-5\right)\left(3x+1+\dfrac{1}{\sqrt{6-x}+1}+\dfrac{3}{4+\sqrt{3x+1}}\right)=0\)
\(\Leftrightarrow x=5\left(\text{vì với }x\ge-\dfrac{1}{3}\text{thì }3x+1+\dfrac{1}{\sqrt{6-x}+1}+\dfrac{3}{4+\sqrt{3x+1}}>0\right)\)
Tập nghiệm phương trình \(S=\left\{5\right\}\)
c) ĐKXĐ : \(x\ge1\)
\(x^2-1=2\sqrt{2x+1}\Leftrightarrow x^2+2x+1-\left(2x+1\right)-2\sqrt{2x+1}-1=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(\sqrt{2x+1}+1\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+1}\right)\left(x+\sqrt{2x+1}+2\right)=0\)
\(\Leftrightarrow x=\sqrt{2x+1}\left(\text{vì }x+\sqrt{2x+1}+2>0\text{ với }x\ge1\right)\)
Khi \(x=\sqrt{2x+1}\Leftrightarrow\left\{{}\begin{matrix}x^2-2x-1=0\\x\ge1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=2\\x\ge1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}+1\\x\ge1\end{matrix}\right.\Leftrightarrow x=\sqrt{2}+1\)
