Ta có: \(A=a^3+b^3+ab\)

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

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

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

Mà a+b=1 =>b=1-a

\(\Rightarrow A=a^2+\left(1-a\right)^2\)

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

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

\(=2\left(a^2-a+\frac{1}{2}\right)\)

\(=2\left(a^2-a+\frac{1}{4}\right)+\frac{1}{2}\)

\(=2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\)

Ta có : \(2\left(a-\frac{1}{2}\right)^2\ge0\forall a\)

\(\Rightarrow2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall a\)

Hay: \(A\ge\frac{1}{2}\forall a\)

Dấu = xảy ra khi : \(2\left(a-\frac{1}{2}\right)^2=0\Rightarrow a=\frac{1}{2}\)

\(\Rightarrow b=1-\frac{1}{2}=\frac{1}{2}\)

Vậy MinA =1/2 tại a=b=1/2

Đây mà toán lớp 9 

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{ac+bc+c^2}\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)=-\left(a+b\right)ab\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+\left(a+b\right)ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

<=> a +  b = 0 hoặc b + c = 0 hoặc c + a = 0

<=> a = -b hoặc b = -c hoặc c = -a

Vậy...

Ta có : 

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Rightarrow\frac{bc+ca+ab}{abc}=\frac{1}{a+b+c}\)

\(\Rightarrow\left(bc+ca+ab\right)\left(a+b+c\right)=abc\)

\(\Rightarrow\left(bc+ac+ab\right)\left(a+b+c\right)-abc=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow\hept{\begin{cases}a+b=0\\b+c=0\\c+a=0\end{cases}}\)

a) \(^{x^4-y^4}\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=\left[\left(x-y\right).\left(x+y\right)\right].\left(x^2-y^2\right)\)

\(=\left(x-y\right).\left(x+y\right).\left(x^2-y^2\right)\)

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

\(=\left[\left(3x-2y\right)+\left(2x-3y\right)\right].\left[\left(3x-2y\right)-\left(2x-3y\right)\right]\)

\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)

b) \(x^2-3y^2\)

\(=\left(x-3y\right)\left(x+3y\right)\)

d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

\(=9\left(x-y\right)^2+4\left(x-y\right)^2\)

\(=\left(x-y\right).\left(9+4\right)\)

\(=\left(x-y\right).13\)

\(=13\left(x-y\right)\)

f) \(x^3+27\)

\(=x^3+3^3\)

\(=\left(x+3\right)\left(x^2-x.3+3^2\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

\(=\left(5x-1\right)\left(5x^2+5x.1+1^2\right)\)

\(=\left(5x-1\right)\left(5x^2+5x+1\right)\)

\(a,x^4-y^4=\left(x^2+y^2\right)\left(x^2-y^2\right)=\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)

\(b,x^2-3y^2=\left(x+\sqrt{3}y\right)\left(x-\sqrt{3}y\right)\)

cn lại tg tự nha bn

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=-\frac{1}{c^3}\)

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(\frac{1}{a}+\frac{1}{b}\right)=0\)

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(\frac{1}{-c}\right)=0\)

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=0\)

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

\(abc.\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=3\)

\(\Rightarrow A=3\)

Vậy \(A=3\)

Tham khảo nhé~