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b. 2 + \(\sqrt{2x-1}=x\) ĐKXĐ: \(x\ge0,5\)
<=> \(\sqrt{2x-1}\) = x - 2
<=> 2x - 1 = (x - 2)2
<=> 2x - 1 = x2 - 4x + 4
<=> -x2 + 2x + 4x - 4 - 1 = 0
<=> -x2 + 6x - 5 = 0
<=> -x2 + 5x + x - 5 = 0
<=> -(-x2 + 5x + x - 5) = 0
<=> x2 - 5x - x + 5 = 0
<=> x(x - 5) - (x - 5) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)
\(\Leftrightarrow16-3\left(x+1\right)< 24+2\left(x-1\right)\)
=>16-3x-3<24+2x-2
=>-3x+13<2x+22
=>-5x<9
hay x>-9/5
|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
\(đk:\left\{{}\begin{matrix}\Delta\ge0\\0< x1\le x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5^2-4\left(-m^2+m+6\right)\ge0\\\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-4m+1=\left(2m-1\right)^2\ge0\left(đúng\right)\\\left\{{}\begin{matrix}5>0đúng\\-m^2+m+6>0\Leftrightarrow-2< m< 3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-2< m< 3\)
\(\Rightarrow\dfrac{1}{\sqrt{x1}}+\dfrac{1}{\sqrt{x2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{\sqrt{x1}+\sqrt{x2}}{\sqrt{x1x2}}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x1+x2+2\sqrt{x1x2}}{x1x2}=\dfrac{9}{4}\Leftrightarrow\dfrac{5+2\sqrt{-m^2+m+6}}{-m^2+m+6}=\dfrac{9}{4}\)
\(đặt::\sqrt{-m^2+m+6}=t\ge0\Rightarrow\dfrac{5+2t}{t^2}=\dfrac{9}{4}\)
\(\Rightarrow9t^2-8t-20=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{10}{9}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-m^2+m+6}=2\Leftrightarrow\left[{}\begin{matrix}m=2\left(tm\right)\\m=-1\left(tm\right)\end{matrix}\right.\)
Viết nhầm biến thôi :v. Sửa''ss
\(\dfrac{\left(x+y\right)^2}{4}+\dfrac{x+y}{2}\ge x\sqrt{y}+y\sqrt{x}\)
Ta có: \(VT\ge\dfrac{4xy}{4}+\dfrac{x}{2}+\dfrac{x}{2}=\dfrac{xy}{2}+\dfrac{xy}{2}+\dfrac{x}{2}+\dfrac{y}{2}\)
\(\Leftrightarrow VT\ge2\sqrt{\dfrac{xy}{2}.\dfrac{x}{2}}+2\sqrt{\dfrac{xy}{2}.\dfrac{y}{2}}\)
\(\Leftrightarrow VT\ge x\sqrt{y}+y\sqrt{x}\)(đpcm)
Đẳng thức xảy ra khi \(\left[{}\begin{matrix}x=y=1\\x=y=0\end{matrix}\right.\)
