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Lời giải:
ĐKXĐ: \(2\leq x\leq 7\)
PT \(\Leftrightarrow (x^2-3x)+(1-\sqrt{x-2})+(2-\sqrt{7-x})=0\)
\(\Leftrightarrow x(x-3)-\frac{x-3}{\sqrt{x-2}+1}+\frac{x-3}{\sqrt{7-x}+2}=0\)
\(\Leftrightarrow (x-3)\left[x-\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{7-x}+2}\right]=0\)
Ta thấy: \(x\geq 2>1; \sqrt{x-2}+1\geq 1\Rightarrow \frac{1}{\sqrt{x-2}+1}\leq 1; \frac{1}{\sqrt{7-x}+2}>0\)
\(\Rightarrow x-\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{7-x}+2}>0\)
\(\Rightarrow x-\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{7-x}+2}\neq 0\)
Do đó: \(x-3=0\Leftrightarrow x=3\) (thỏa mãn)
Vậy PT có nghiệm $x=3$
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\(x^2-3x+1=\dfrac{5\sqrt{3}}{3}\sqrt{x^4+x^2+1}\)
\(\Leftrightarrow\)\(\left(x^2-3x+1\right)^2=\dfrac{25}{3}\left(x^4+x^2+1\right)\)
\(\Leftrightarrow\)\(x^4-6x^3+11x^2-6x+1=\dfrac{25}{3}x^4+\dfrac{25}{3}x^2+\dfrac{25}{3}\)
\(\Leftrightarrow11x^4+9x^3-4x^2+9x+11=0\)
\(\Leftrightarrow\left(x+1\right)\left(11x^3-2x^2-2x+11\right)=0\)
\(\Rightarrow x=-1\)
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\(Đkxđ:x\ge0\)
Ta có: Bất phương trình tương đương với:
\(\left(1+\sqrt{x}\right)\left(\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\right)=2\)
Áp dụng BĐT Cô - si ta có:
\(\frac{1}{\sqrt{3x+1}}=\sqrt{\frac{1}{x+1}.\frac{x+1}{3x+1}}\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{x+1}{3x+1}\right)\)
\(\sqrt{\frac{x}{3x+1}}=\sqrt{\frac{1}{2}.\frac{2x}{3x+1}}\le\frac{1}{2}\left(\frac{1}{2}+\frac{2x}{3x+1}\right)\)
\(\Rightarrow\frac{1+\sqrt{x}}{\sqrt{3x+1}}\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{2}+1\right)=\frac{1}{2}\left(\frac{1}{x+1}+\frac{3}{2}\right)\left(1\right)\)
\(\frac{1}{\sqrt{x+3}}=\sqrt{\frac{1}{2}.\frac{2}{x+3}}\le\frac{1}{2}\left(\frac{1}{2}+\frac{2}{x+3}\right)\)
\(\frac{\sqrt{x}}{\sqrt{x+3}}=\sqrt{\frac{x}{x+1}.\frac{x+1}{x+3}}\le\frac{1}{2}\left(\frac{x}{x+1}+\frac{x+1}{x+3}\right)\)
\(\Rightarrow\frac{1+\sqrt{x}}{\sqrt{x+3}}\le\frac{1}{2}\left(\frac{x}{x+1}+\frac{3}{2}\right)\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow\left(1+\sqrt{x}\right)\left(\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\right)\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{x}{x+1}+3\right)=2\)
Đẳng thức xảy ra \(\Leftrightarrow x=1\)
Vậy nghiệm của pt là \(x=1\)
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a: \(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{2x+1}-2\right)=0\)
=>2x-1=0 hoặc 2x+1=4
=>2x=1 hoặc 2x=3
=>x=3/2 hoặc x=1/2
b: \(\Leftrightarrow3x+2=2\left(x+2\right)\)
=>3x+2=2x+4
=>x=2(nhận)
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dkxd: -2 ≤ x ≤ 2
\(x+\sqrt{4-x^2}=2+3x\sqrt{4-x^2}\)
\(\Leftrightarrow x-2=3x\sqrt{4-x^2}-\sqrt{4-x^2}\)
\(\Leftrightarrow x-2=\sqrt{4-x^2}\left(3x-1\right)\)
\(\Leftrightarrow x^2-4x+4=\left(4-x^2\right)\left(9x^2-6x+1\right)\)
\(\Leftrightarrow x^2-4x+4=-9x^4+6x^3+35x^2-24x+4\)
\(\Leftrightarrow9x^4-6x^3-34x^2+20x=0\)
\(\Leftrightarrow x\left(9x^3-6x^2-34x+20\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(9x^2+12x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\9x^2+12x-10=0\end{matrix}\right.\)
+) => x= 0 (nhận)
+) x - 2 = 0 => x = 2 (nhận)
+) \(9x^2+12x-10=0\)
\(\Leftrightarrow\left(9x^2+12x+4\right)-14=0\)
\(\Leftrightarrow\left(3x+2\right)^2=14\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=\sqrt{14}\\3x+2=-\sqrt{14}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{14}}{3}\left(loai\right)\\x=\dfrac{-2-\sqrt{14}}{3}\left(nhan\right)\end{matrix}\right.\)
Vậy pt có 3 nghiệm ...........