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1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
Bài 1:
\(M=\dfrac{9}{\sqrt{11}-\sqrt{2}}-\dfrac{\sqrt{22}-\sqrt{10}}{\sqrt{11}-\sqrt{5}}-\dfrac{22}{\sqrt{11}}\)
\(=\dfrac{9\left(\sqrt{11}+\sqrt{2}\right)}{11-2}-\dfrac{\sqrt{2}\left(\sqrt{11}-\sqrt{5}\right)\left(\sqrt{11}+\sqrt{5}\right)}{11-5}-\dfrac{2.\left(\sqrt{11}\right)^2}{\sqrt{11}}\)
\(=\sqrt{11}+\sqrt{2}-\sqrt{2}-2\sqrt{11}=-\sqrt{11}\)
\(M=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}+\dfrac{a-b}{\sqrt{a}+\sqrt{b}}+\dfrac{2b}{\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}+\dfrac{2\left(\sqrt{b}\right)^2}{\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}+\sqrt{a}-\sqrt{b}+2\sqrt{b}=2\sqrt{a}\)
Bài 2:
a)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)+\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2}{\sqrt{x}+1}\) (*)
b)
Thay x = 0,25 vào (*), ta có:
\(M=\dfrac{2}{\sqrt{\dfrac{1}{4}}+1}=\dfrac{4}{3}\)
c)
\(M\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\ge1\)
\(\Leftrightarrow2\ge\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}\le1\)
\(\Leftrightarrow x\le1\)
mà x khác 1 và x > 0(theo ĐKXĐ)
=> 0 < x < 1 thì M \(\ge\) 1
\(x=9\Rightarrow\sqrt{x}=3\Rightarrow A=\frac{3+2}{3-5}=\frac{5}{-2}=-\frac{5}{2}\\ \)
\(B=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{x-25}=\frac{3.\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}+\frac{20-2\sqrt{x}}{\left(x+\sqrt{5}\right).\left(x-\sqrt{5}\right)}\)
\(=\frac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}+5}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{1}{\sqrt{x}-5}\)
\(A=B.\left|x-4\right|\Leftrightarrow\left|x-4\right|=A:B=\frac{\sqrt{x}+2}{\sqrt{x}-5}:\frac{1}{\sqrt{x}-5}=\sqrt{x}+2\)
\(\Rightarrow\left(x-4\right)^2=\left(\sqrt{x}+2\right)^2\Leftrightarrow x^2-8x+16=x+4\sqrt{x}+4\)
\(\Leftrightarrow x^2-9x-4\sqrt{x}+12=0\Leftrightarrow x.\left(x-9\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x.\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x\sqrt{x}+3x-4\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\left(x\sqrt{x}-x\right)+\left(4x-4\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x.\left(\sqrt{x}-1\right)+4.\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(x+4\sqrt{x}+4\right)=0\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-3=0\\\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}}\)(Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\left(\sqrt{x}+2\right)^2\ge4>0\))
Lời giải:
a.
Áp dụng BĐT Bunhiacopxky:
$A^2=(\sqrt{x-1}+\sqrt{9-x})^2\leq (x-1+9-x)(1+1)=16$
$\Rightarrow A\leq 4$
Vậy $A_{\max}=4$. Giá trị này đạt tại $x=5$
b.
$A=\frac{3(\sqrt{x}+2)+5}{\sqrt{x}+2}=3+\frac{5}{\sqrt{x}+2}$
Để $A$ nguyên thì $\frac{5}{\sqrt{x}+2}=m$ với $m$ nguyên dương
$\Leftrightarrow \sqrt{x}+2=\frac{5}{m}$
$\sqrt{x}=\frac{5-2m}{m}$
Vì $\sqrt{x}\geq 0$ nên $\frac{5-2m}{m}\geq 0$
Mà $m$ nguyên dương nên $5-2m\geq 0$
$\Leftrightarrow m\leq 2,5$.
$\Rightarrow m=1; 2$
$\Rightarrow x=9; x=\frac{1}{4}$