Áp dụng BĐT \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\left(x,y,z>0\right)\) ta có :

\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)

\(\ge\frac{\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)^2}{3}\ge\frac{\left(a+b+c+\frac{9}{a+b+c}\right)^2}{3}=\frac{\left(1+9\right)^2}{3}=\frac{100}{3}\)

Vậy \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{100}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

\(\left(6x^2-5\right)\left(2x+3\right)=12x^3+18x^2-10x-15\)

\(\left(6x^2-5\right)\left(2x+3\right)=12x^3+18x^2-10x-15\)

uyaa,f đăng nhầm rr

Ta có : \(\left(x^4-2x^3+2x+a\right)\div\left(x-1\right)^2\)

\(\Rightarrow\left(x^4-2x^3+2x+a\right)\div\left(x^2-2x+1\right)\)

Ta đặt phép chia :

x^4 - 2x^3 + 2x + a x^2 - 2x + 1 x^2 - 1 x^4 - 2x^3 + x^2 -x^2 + 2x + a -x^2 + 2x - 1 a + 1

Để đa thức trên chia hết thì \(a+1=0\)

                                          \(\Rightarrow a=-1\)

Vậy \(a=-1\) thì \(\left(x^4-2x^3+2x+a\right)⋮\left(x-1\right)^2\)

thiếu dữ kiện a,b,c khác 0 nha

2(x + 1)(y + 1) = (x + y)(x + y + 2)

2(xy + y + x + 1) = x² + 2xy +y² + 2x + 2y

2xy + 2y + 2x + 2 = 2xy + 2y + 2x + x² + y²

2xy + 2y + 2x + x² + y² = 2xy + 2y + 2x + x² + y²

=> 2(x+1)(y+1)=(x+y)(x+y+2)