\(a^3+b^3+a^2c+b^2c-abc=a^2\left(a+b+c\right)+bc\left(b-a\right)=bc\left(b-a\right)\)

Do \(0\le a,b,c\le1\)

nên\(\left\{{}\begin{matrix}\left(a^2-1\right)\left(b-1\right)\ge0\\\left(b^2-1\right)\left(c-1\right)\ge0\\\left(c^2-1\right)\left(a-1\right)\ge0\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}a^2b-b-a^2+1\ge0\\b^2c-c-b^2+1\ge0\\c^2a-a-c^2+1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2b\ge a^2+b-1\\b^2c\ge b^2+c-1\\c^2a\ge c^2+a-1\end{matrix}\right.\)

Ta cũng có:

\(2\left(a^3+b^3+c^3\right)\le a^2+b+b^2+c+c^2+a\)

Do đó \(T=2\left(a^3+b^3+c^3\right)-\left(a^2b+b^2c+c^2a\right)\)

\(\le a^2+b+b^2+c+c^2+a\)\(-\left(a^2+b-1+b^2+c-1+c^2+a-1\right)\)

\(=3\)

Vậy GTLN của T=3, đạt được chẳng hạn khi \(a=1;b=0;c=1\)

giúp mình câu hỏi này với ah.

`a)a(2+b)+b(a+2)`

`=2a+ab+ab+2b`

`=2(a+b)+2ab`

`=2.10+2.(-36)`

`=20-72=-52`

`b)a^2+b^2`

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

`=10^2-2.(-36)`

`=100+72=172`

`c)a^3+b^3`

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

`=10[(a+b)^2-3ab]`

`=10[10^2-3.(-36)]`

`=10(100+108)`

`=10.208=2080`

a, \(=>2a+ab+ab+2b=2\left(a+b+ab\right)=2\left(10-36\right)=-52\)

b, \(a^2+b^2=a^2+2ab+b^2-2ab=\left(a+b\right)^2-2ab=\left(10\right)^2-2\left(-36\right)=172\)

c, \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=10\left[\left(a+b\right)^2-3ab\right]\)

\(=10\left[10^2-3\left(-36\right)\right]=2080\)

\(a\left(2+b\right)+b\left(a+2\right)=2ab+2\left(a+b\right)=2.\left(-36\right)+2.10=-52\)

\(a^2+b^2=\left(a+b\right)^2-2ab=172\)

\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=2080\)

b) Ta có: \(a^2+b^2\)

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

\(=3^2+2\cdot\left(-2\right)=9-4=5\)

c) Ta có: \(a^3-b^3\)

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

\(=3^3-3\cdot\left(-2\right)\cdot3\)

\(=27+18=45\)

cho mình hỏi yêu cầu đề bài là gì vậy?

Ta có hằng đẳng thức: 

\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc=0.\left(a^2+b^2+c^2-ab-bc-ca\right)+3.1=0+3=3\)

\(5,M=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\\ M=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\\ M=1\left(1-3ab\right)=1-3ab\ge1-\dfrac{3\left(a+b\right)^2}{4}=1-\dfrac{3}{4}=\dfrac{1}{4}\\ M_{min}=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 5:

\(a+b=1\Rightarrow a=1-b\)

\(M=a^3+b^3=\left(1-b\right)^3+b^3=1-3b+3b^2-b^3+b^3\)

\(=1-3b+3b^2=3\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(minM=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 7:

\(a^3+b^3+abc\ge ab\left(a+b+c\right)\)

\(\Leftrightarrow a^3+b^3+abc-ab\left(a+b+c\right)\ge0\)

\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng do a,b dương)

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

giúp mình vs

5.

Với mọi a;b ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow2a^2+2b^2\ge a^2+b^2+2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)

\(M=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=a^2+b^2-ab\)

\(M=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\left(a+b\right)^2=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\ge\dfrac{3}{2}.\dfrac{1}{2}-\dfrac{1}{2}=\dfrac{1}{4}\)

\(M_{min}=\dfrac{1}{4}\) khi \(a=b=\dfrac{1}{2}\)

6.

Do \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=2>0\)

Mà \(a^2-ab+b^2>0\Rightarrow a+b>0\)

Mặt khác với mọi a;b ta có:

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2+b^2+2ab\ge4ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow ab\le\dfrac{1}{4}\left(a+b\right)^2\) \(\Rightarrow-ab\ge-\dfrac{1}{4}\left(a+b\right)^2\)

Từ đó:

\(2=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\ge\left(a+b\right)^3-3.\dfrac{1}{4}\left(a+b\right)^2\left(a+b\right)=\dfrac{1}{4}\left(a+b\right)^3\)

\(\Rightarrow\left(a+b\right)^3\le8\Rightarrow a+b\le2\)

\(N_{max}=2\) khi \(a=b=1\)

7.

Ta có:

\(a^3+b^3+abc=\left(a+b\right)\left(a^2+b^2-ab\right)+abc\ge\left(a+b\right)\left(2ab-ab\right)+abc\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b\)

8.

\(\left|a+b\right|>\left|a-b\right|\Leftrightarrow\left(a+b\right)^2>\left(a-b\right)^2\)

\(\Leftrightarrow a^2+2ab+b^2>a^2-2ab+b^2\)

\(\Leftrightarrow4ab>0\Leftrightarrow ab>0\)

\(\Rightarrow a;b\) cùng dấu