Mk ms tìm được GTNN thôi!

Ta có: A = a³ + b³ = (a + b)(a² + b² - ab) = (a + b)(1 - ab)

Áp dụng BĐT Cô-si cho 2 số ko âm a² và b² ta có:

a² + b² \(\ge\) 2ab

\(\Leftrightarrow\) 1 \(\ge\) 2ab

\(\Leftrightarrow\) 1 - 2ab \(\ge\) 0

\(\Leftrightarrow\) 1 - ab \(\ge\) ab

\(\Rightarrow\) A \(\ge\) ab(a + b)

Dấu "=" xảy ra khi và chỉ khi a = b = \(\sqrt{0,5}\)

\(\Rightarrow\) A \(\ge\) 0,5 . 2\(\sqrt{0,5}\) = \(\sqrt{0,5}\)

Vậy ...

Chúc bn học tốt!

\(a^2+b^2=1\Rightarrow\left\{{}\begin{matrix}0\le a\le1\\0\le b\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3\le a^2\\b^3\le b^2\end{matrix}\right.\)

\(\Rightarrow a^3+b^3\le a^2+b^2=1\)

\(A_{max}=1\) khi \(\left(a;b\right)=\left(0;1\right);\left(1;0\right)\)

\(a^3+a^3+\left(\dfrac{1}{\sqrt{2}}\right)^3\ge\dfrac{3}{\sqrt{2}}a^2\)

\(b^3+b^3+\left(\dfrac{1}{\sqrt{2}}\right)^3\ge\dfrac{3}{\sqrt{2}}b^2\)

Cộng vế:

\(2\left(a^3+b^3\right)+\dfrac{\sqrt{2}}{2}\ge\dfrac{3}{\sqrt{2}}\left(a^2+b^2\right)=\dfrac{3\sqrt{2}}{2}\)

\(\Rightarrow a^3+b^3\ge\dfrac{\sqrt{2}}{2}\)

\(A_{min}=\dfrac{\sqrt{2}}{2}\) khi \(a=b=\dfrac{\sqrt{2}}{2}\)

Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

\(a^2+b^2=a^3+b^3=a^4+b^4\)

\(\Rightarrow\left(a^3+b^3\right)^2=\left(a^2+b^2\right)\left(a^4+b^4\right)\)

\(\Rightarrow a^6+b^6+2a^3b^3=a^6+b^6+a^2b^4+a^4b^2\)

\(\Rightarrow2a^3b^3=a^2b^2\left(a^2+b^2\right)\)

\(\Rightarrow2ab=a^2+b^2\)

\(\Rightarrow\left(a-b\right)^2=0\)

\(\Rightarrow a=b\)

Thế vào \(a^2+b^2=a^3+b^3\)

\(\Rightarrow a^2+a^2=a^3+a^3\Rightarrow2a^3=2a^2\Rightarrow a=b=1\)

\(\Rightarrow a+b=2\)

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

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

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

a: \(a^2+4a=b^2+4b+1\)

=>\(a^2+4a-b^2-4b=0\)

=>(a-b)(a+b)+4(a-b)=0

=>(a-b)(a+b+4)=0

mà a-b<>0

nên a+b+4=0

=>a+b=-4

b: Đặt \(X=a^3+b^3\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=\left(-4\right)^3-3ab\cdot\left(-4\right)=-64+12ab\)

\(a^2+4a=1\)

=>\(a^2+4a-1=0\)

=>\(a^2+4a+4-5=0\)

=>\(\left(a+2\right)^2=5\)

=>\(\left[\begin{array}{l}a+2=\sqrt5\\ a+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\sqrt5-2\\ a=-\sqrt5-2\end{array}\right.\)

\(b^2+4b=1\)

=>\(b^2+4b-1=0\)

=>\(b^2+4b+4-5=0\)

=>\(\left(b+2\right)^2=5\)

=>\(\left[\begin{array}{l}b+2=\sqrt5\\ b+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}b=\sqrt5-2\\ b=-\sqrt5-2\end{array}\right.\)

Vì a<>b nên sẽ có hai trường hợp sau:

TH1: \(a=\sqrt5-2;b=-\sqrt5-2\)

=>\(ab=\left(\sqrt5-2\right)\left(-\sqrt5-2\right)=-\left(\sqrt5-2\right)\left(\sqrt5+2\right)=-1\)

X=-64+12ab

=-64-12

=-76

TH2: \(a=-\sqrt5-2;b=\sqrt5-2\)

=>\(ab=\left(\sqrt5-2\right)\left(-\sqrt5-2\right)=-\left(\sqrt5-2\right)\left(\sqrt5+2\right)=-1\)

X=-64+12ab

=-64-12

=-76

Vậy: X=-76

c: Đặt \(Y=a^4+b^4\)

\(=\left(a^2+b^2\right)^2-2a^2b^2\)

\(=\left\lbrack\left(a+b\right)^2-2ab\right\rbrack^2-2\cdot\left(ab\right)^2\)

\(=\left\lbrack\left(-4\right)^2-2\cdot\left(-1\right)\right\rbrack^2-2\cdot\left(-1\right)^2=\left\lbrack16+2\right\rbrack^2-2\)

\(=18^2-2\)

=324-2

=322

VP `=(a+b)(a^2-ab+b^2)`

`=a^3-a^2b+ab^2+a^2b-ab^2+b^3`

`=a^3+(a^2b-a^2b)+(ab^2-ab^2)+b^3`

`=a^3+b^3`

.

VP `=(a-b)(a^2+ab+b^2)`

`=a^3+a^2b+ab^2-a^2b-ab^2-b^3`

`=a^3+(a^2b-a^2b)+(ab^2-ab^2)-b^3`

`=a^3-b^3`

đúng rồi mà

Hướng dẫn: Chọn đáp án B