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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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a/
ĐK \(x^2-6x+6\ge0\)
\(\text{pt }\Leftrightarrow\left(x^2-6x+6\right)-4\sqrt{x^2-6x+6}+3=0\)
Đặt \(t=\sqrt{x^2-6x+6};t\ge0\)
pt thành \(t^2-4t+3=0\Leftrightarrow t=3\text{ hoặc }t=1\)
\(+t=1\Rightarrow x^2-6x+6=1^2\Leftrightarrow x^2-6x+7=0\Leftrightarrow t=3+\sqrt{2}\text{ hoặc }t=3-\sqrt{2}\)
\(+t=3\Rightarrow x^2-6x+6=3^2\Leftrightarrow x^2-6x-3=0\Leftrightarrow x=3+2\sqrt{3}\text{ hoặc }x=3-2\sqrt{3}\)
Vậy ....
b/
ĐK: \(x^2+3x\ge0\)
\(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\Leftrightarrow-\left(x^2+3x\right)-3\sqrt{x^2+3x}+10=0\)
\(\Leftrightarrow\left(\sqrt{x^2+3x}-2\right)\left(\sqrt{x^2+3x}+5\right)=0\)
\(\Leftrightarrow\sqrt{x^2+3x}=2\text{ hoặc }\sqrt{x^2+3x}=-5\text{ (loại)}\)
\(\Leftrightarrow x^2+3x-2^2=0\Leftrightarrow x=1\text{ hoặc }x=-4\)
Vậy ....
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\(đk:2\le x\le4\) \(pt\Leftrightarrow\sqrt{x-2}+\sqrt{4-x}=x-2\sqrt{3x}+5\)
\(\left(\sqrt{x-2}+\sqrt{4-x}\right)^2\le2\left(x-2+4-x\right)=4\Rightarrow\sqrt{x-2}+\sqrt{4-x}\le2\)
\(x-2\sqrt{3x}+5=\sqrt{x}^2-2\sqrt{3x}+5=\sqrt{x}^2-2\sqrt{3x}+3+2=\left(\sqrt{x}-\sqrt{3}\right)^2+2\ge2\)
\(\Rightarrow\left\{{}\begin{matrix}VT\le2\\VP\ge2\end{matrix}\right.\) dấu"=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{4-x}=2\\\left(\sqrt{x}-\sqrt{3}\right)^2+2=2\end{matrix}\right.\)
\(\Leftrightarrow x=3\left(tm\right)\)
(ủa đề sai chỗ nào ta?)
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 ...........