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\(M=a^2+ab+b^2-3a-3b+2001\)
\(\Rightarrow2M=2a^2+2ab+2b^2-6a-6b+4002\)
\(=\left[\left(a+b\right)^2-2\left(a+b\right).2+4\right]+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+3996\)
\(=\left(a+b-2\right)^2+\left(a-1\right)^2+\left(b-1\right)^2+3996\ge3996\)
\(\Rightarrow M\ge1998\)
\(minM=1998\Leftrightarrow a=b=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\sqrt{1-x}+\sqrt{1+x}\) \(\left(-1\le x\le1\right)\)
\(=1.\sqrt{1-x}+1.\sqrt{1+x}\)
Áp dụng BĐT Bunhiacopxki, ta có:
\(A=1.\sqrt{1-x}+1.\sqrt{1+x}\)
\(\le\sqrt{\left(1^2+1^2\right).\left(1-x+1+x\right)}=\sqrt{2.2}=2\)
Vậy \(A_{max}=2\), đạt được khi và chỉ khi \(\dfrac{1}{\sqrt{1-x}}=\dfrac{1}{\sqrt{1+x}}\Leftrightarrow1-x=1+x\Leftrightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=a^2+ab+b^2-3a-3b+2001\)
\(\Rightarrow2M=2a^2+2ab+2b^2-6a-6b+4002\)
\(=\left(a^2+2ab+b^2\right)-4\left(a+b\right)+4+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+3996\)
\(=\left(a+b-2\right)^2+\left(a-1\right)^2+\left(b-1\right)^2+3996\ge3996\)
\(\Rightarrow M\ge1998\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge0\)
\(A=\dfrac{3\sqrt{x}}{\sqrt{x}+2}=\dfrac{3\left(\sqrt{x}+2\right)}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}+2}=3-\dfrac{6}{\sqrt{x}+2}\)
Do \(\sqrt{x}+2\ge2\Leftrightarrow\dfrac{6}{\sqrt{x}+2}\le\dfrac{6}{2}=3\)
\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+2}\ge-3\)
\(A=3-\dfrac{6}{\sqrt{x}+2}\ge3-3=0\)
\(minA=0\Leftrightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=a+\sqrt{a}\)
\(=\left[\left(\sqrt{a}\right)^2+2.\sqrt{a}.\dfrac{1}{2}+\dfrac{1}{4}\right]-\dfrac{1}{4}\)
\(=\left(\sqrt{a}+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=-\dfrac{1}{4}+\left(\sqrt{a}+\dfrac{1}{2}\right)^2\)
Vì \(\left(\sqrt{a}+\dfrac{1}{2}\right)^2\) ≥ 0
⇒ M≤ \(-\dfrac{1}{4}\)
Min M=\(-\dfrac{1}{4}\)
ĐKXĐ: \(a\ge0\)
Khi đó ta có: \(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ge0\end{matrix}\right.\) \(\Rightarrow a+\sqrt{a}\ge0\)
\(M_{min}=0\) khi \(a=0\)
A = sina^6 + cosa^6 =( sina^2)^3 + (cosa^2)^3 = (sina^2 + cosa^2)(sina^4 + cosa^4 - sina^2cosa^2)
= (sina^2 + cosa^2)^2 - 3sina^2cosa^2 = 1 - 3sina^2 (1-sina^2) = 3sina^4 - 3sina^2 + 1 =
= 3( sina^4 - sina^2 + 1/4) + 1/4 = 3(sina^2 - 1/2)^2 + 1/4 \(\ge\frac{1}{4}\) dấu "=" xảy ra khi và chỉ khi
sina^2 = 1/2 khi và chỉ khi a = .......
suy ra Amin = 1/4