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a- \(\sqrt{a}\) + 1 = a - \(\sqrt{a}\) + 1/4 +3/4 = (\(\sqrt{a}\)-1/2)^2 +3/4 > 3/4
Dấu đẳng thức xảy ra <=> a=1/4
Vậy Min = 3/4
Ta có : \(a-\sqrt{a}+1=a-2.\frac{1}{2}\sqrt{a}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2+\frac{3}{4}>=\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=\frac{1}{4}\)
Vậy Min \(=\frac{3}{4}\)đạt tại \(x=\frac{1}{4}\)
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Lời giải:
a.
\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-(2\sqrt{x}+1)+2(\sqrt{x}+1)\)
\(=\sqrt{x}(\sqrt{x}-1)-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)
b.
$A=x-\sqrt{x}+1=(x-\sqrt{x}+\frac{1}{4})+\frac{3}{4}$
$=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}\geq 0+\frac{3}{4}=\frac{3}{4}$
$\Rightarrow A_{\min}=\frac{3}{4}$
Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$
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\(\sqrt{2}A=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(a+1\right)}\le\frac{2a+2b+a+b+2}{2}=\frac{8}{2}=4\)
\(\Rightarrow A\le\frac{4}{\sqrt{2}}=2\sqrt{2}.\text{Dấu "=" xảy ra khi:}a=b=1\)
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\(A=1-|1-3x|+|3x-1|^2\)
\(=\left(|3x-1|-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow minA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)hoặc \(x=\frac{1}{6}\)
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a: \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Ta có: \(a-\sqrt{a}+1=a-2.\frac{1}{2}\sqrt{a}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2+\frac{3}{4}\) >= 3/4
Dấu "=" xảy ra khi a=1/4
Vậy min=3/4 đạt tại x=1/4