a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)

Ta có B = (3a+b)(a - 2b)

Ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

\(\Rightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}\Leftrightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(a+b\right)^2\ge4ab\left(1\right)\\\left(a+b\right)^2\le2\left(a^2+b^2\right)\left(2\right)\end{cases}}\)

Theo đề bài:

\(a+b+3ab=1\)

\(\Leftrightarrow4\left(a+b\right)+12ab=4\)

\(\Leftrightarrow4\left(a+b\right)+3\left(a+b\right)^2\ge4\left(theo\left(1\right)\right)\)

\(\Leftrightarrow3\left(a+b\right)^2+4\left(a+b\right)-4\ge0\)

\(\Leftrightarrow\left(a+b+2\right)\left[3\left(a+b\right)-2\right]\ge0\)

\(\Leftrightarrow3\left(a+b\right)-2\ge0\left(a,b>0\Rightarrow a+b+2>0\right)\)

\(\Leftrightarrow a+b\ge\frac{2}{3}\)

`\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\ge\frac{4}{9}\left(theo\left(2\right)\right)\)

Áp dụng các kết quả trên, ta có:

\(\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^2\le2\left(1-a^2+1-b^2\right)\)\(=4-2\left(a^2+b^2\right)\le4-\frac{4}{9}=\frac{32}{9}\)

\(\Rightarrow\sqrt{1-a^2}+\sqrt{1-b^2}\le\frac{4\sqrt{2}}{3}\)

Ta có: \(\frac{3ab}{a+b}=\frac{1-\left(a+b\right)}{a+b}=\frac{1}{a+b}-1\le\frac{1}{\frac{2}{3}}-1=\frac{1}{2}\)

\(\Rightarrow A\le\frac{4\sqrt{2}}{3}+\frac{1}{2}\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b+3ab=1\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\3a^2+2a-1=0\end{cases}\Leftrightarrow}a=b=\frac{1}{3}\left(a,b>0\right)}\)

Vậy max A là \(\frac{4\sqrt{2}}{3}+\frac{1}{2}\Leftrightarrow a=b=\frac{1}{3}\)

Đáp án D

Đáp án: D.

Đáp án D

thôi mk tự lm đc rồi:

(a^3- 3ab^2)^2=361

=a^6- 6a^4b^2+ 9a^2 b^4

(b^3-3a^2b)^2=9604

=b^6- 6a^2b^4+9a^4 b^2

    cộng 2 vế->(a^2+b^2)^3= 9604+361= 9965

mn check hộ mk nha