^{A = x^100 - 21x^99 - 21x^98 - 21x^97 -...-21x^2 - 21x +2010}

^{A=x^100 - 22x^99 + x^99 -22x^98 + x^98 - ... - 22x +x +2010}

^{A=x^99 (x-22) + x^98 (x-22) + x^97(x-22) + ... + x(x-22) + x +2010}

^{A=(x-22) (x^99 + x^98 + x^97 + ... + x) + x + 2010}

Thay x = 22 vào A, tao có:

A= (22-22) (22^99 + 22^98 + ... +22) + 22 + 2010

A = 0 (22^99 + 22^98 + ... +22) + 2032

A= 0 + 2032

A = 2032

x=22

=>x-1=21

thay 21=x-1 vào A ta được:

A=x¹⁰⁰-(x-1)x⁹⁹-(x-1)x⁹⁸-(x-1)x⁹⁷-...-(x-1)x²-(x-1)x+2010

=x¹⁰⁰-x¹⁰⁰+x⁹⁹-x⁹⁹+x⁹⁸-x⁹⁸+x⁹⁷-...-x³+x²-x²+x+2012

=>A=x+2012

thay x=22 vào A=x+2012 ta được:

A=22+2012=2034

`x^3+y^3+21x+21y`

`=(x+y)(x^2-xy+y^2)+21(x+y)`

`=(x+y)(x^2-xy+y^2+21)`

\(x^3+y^3+21x+21y\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+21\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2+21\right)\)

\(l,x^3+y^3+21x+21y\\=(x^3+y^3)+(21x+21y)\\=(x+y)(x^2-xy+y^2)+21(x+y)\\=(x+y)(x^2-xy+y^2+21)\)

#\(Toru\)

\(1,x^2-16y^2=\left(x-4y\right)\left(x+4y\right)\)

\(2,21x-21y+ax-ay=21\left(x-y\right)+a\left(x-y\right)=\left(21+a\right)\left(x-y\right)\)

\(3,x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x+1\right)^2\)

1) \(21x^2+21y^2+z^2\)

\(=18\left(x^2+y^2\right)+z^2+3\left(x^2+y^2\right)\)

\(\ge9\left(x+y\right)^2+z^2+3.2xy\)

\(\ge2.3\left(x+y\right).z+6xy\)

\(=6\left(xy+yz+zx\right)=6.13=78\)

Dấu "=" xảy ra <=> x = y ; 3(x+y) = z; xy + yz + zx= 13 <=> x = y = 1; z= 6

2) \(x+y+z=3xyz\)

<=> \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3\)

Đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)=> ab + bc + ca = 3

Ta cần chứng minh: \(3a^2+b^2+3c^2\ge6\)

Ta có: \(3a^2+b^2+3c^2=\left(a^2+c^2\right)+2\left(a^2+c^2\right)+b^2\)

\(\ge2ac+\left(a+c\right)^2+b^2\ge2ac+2\left(a+c\right).b=2\left(ac+ab+bc\right)=6\)

Vậy: \(\frac{3}{x^2}+\frac{1}{y^2}+\frac{3}{z^2}\ge6\)

Dấu "=" xảy ra <=> a = c = \(\sqrt{\frac{3}{5}}\); \(b=2\sqrt{\frac{3}{5}}\)

khi đó: \(x=z=\sqrt{\frac{5}{3}};y=\sqrt{\frac{5}{3}}\)