Đặt a+b=x ; b+c=y ; c+a=z, A=(a+b)³.....

Khi đó A= x³+y³+z³-3xyz= (x+y)³- 3xy(x+y) - 3xyz +z³

= (x+y+z)³- 3z(x+y)(x+y+z)- 3xy(x+y+z)

=(x+y+z)(x²+y²+z²+2xy+2yz+2xz-3xz-3zy-3xy)

= (x+y+z)(x²+y²+z²-xy-yz-xz)

tu day em thay vao nhe

Ta có \(\left(a+b+1\right).\left(a^2+b^2\right)+\frac{4}{a+b}\)

\(\ge\left(a+b+1\right).2ab+\frac{4}{a+b}\)

\(=2.\left(a+b\right)+2+\frac{4}{a+b}\)

\(=a+b+2+a+b+\frac{4}{a+b}\)

\(\ge2.\sqrt{a.b}+2+2.\sqrt{\left(a+b\right).\frac{4}{a+b}}=2+2+2\sqrt{4}\)

\(=2+2+4=8\)

Vậy\(\left(a+b+1\right).\left(a^2+b^2\right)+\frac{4}{a+b}\ge8\)với ab=1

\(16x^3-12x^2+3x-7=0\)

\(\Leftrightarrow16x^3-16x^2-3x^2+3x+7x^2-7=0\)

\(\Leftrightarrow16x^2\left(x-1\right)-3x\left(x-1\right)+7\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow16x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\left(7x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(16x^2-3x+7x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(16x^2+4x+7\right)=0\)

<=> x - 1 = 0 

<=> x = 1

\(\Leftrightarrow16x^3-16x^2+4x^2-4x+7x-7=0\)

\(\Leftrightarrow16x^2.\left(x-1\right)+4x.\left(x-1\right)+7.\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right).\left(16x^2+4x+7\right)=0\)

Ta có \(16x^2+4x+7=\left(4x\right)^2+2.4x.\frac{1}{2}+\frac{1}{4}+\frac{27}{4}\)

\(=\left(4x+\frac{1}{2}\right)^2+\frac{27}{4}>0\)

nên \(\left(x-1\right).\left(16x^2+4x+7\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Rightarrow x=1\)

(x^2 + x)^2 - 2.(x^2 + x) - 15

= (x² + x)² - 2(x² + x) + 1 - 16

= (x² + x + 1)² - 16

= (x² + x + 1 - 4)(x² + x + 1 + 4)

= (x² + x - 3)(x² + x + 5)

\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=\left(\left(x^2+x\right)^2-2\left(x^2+x\right)+1\right)-16\)

\(=\left(x^2+x-1\right)^2-4^2\)

\(=\left(x^2+x-1-4\right)\left(x^2+x-1+4\right)\)

\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)