\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2b\)

tương tự cộng theo vế rút gọn ta có đpcm 

1)a)x+y=60

<=>(x+y)^2=3600

<=>x^2+2xy+y^2=3600(1)

mà xy=35 nên 2xy=2.35=70

(1)<=>x^2+70+y^2=3600

<=>x^2+y^2=3530

<=>(x^2+y^2)^2=12460900

<=>x^4+2x^2.y^2+y^4=12460900(2)

mà xy=35 nên 2x.x.y.y=2450

(2)<=>x^4+y^4=123458450

 b)x+y=1

<=>(x+y)^3=1

<=>x^3+3x^2y+3xy^2+y^3=1

<=>x^3+y^3+3xy(x+y)=1

<=>x^3+y^3+3xy=1

=>M=1

x+y=1

<=>x^2+2xy+y^2=1(1)

B=x^3+y^3+3xy(x^2+y^2)+3xy(2xy)

=x^3+y^3+3xy(x^2+2xy+y^2)

=M.1=1(từ(1)

c)

x-y=1

<=>(x-y)^3=1

<=>x^3-3x^2y+3xy^2-y^3=1

<=>x^3-y^3-3xy(x-y)=1

<=>x^3-y^3-3xy=1

=>N=1

giải giúp mình với 

Toán lớp mấy vậy ???\

giúp mình vk mình hơi dốt toán

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

vô nghiệm

\(A=x\left(2x+1\right)-4xy+y\left(2y-1\right)+2000.\)

\(2x^2+x-4xy+2y^2-y+2000\)

\(2\left(x-y\right)^2+x-y+2000=2.25+5+2000\)

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

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

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

\(2\left(x-1\right)\left(x^2+\frac{3}{2}x-2\right)=2\left(x-1\right)\left\{\left(x^2+\frac{2x.3}{4}+\frac{9}{16}\right)-\left(2+\frac{9}{16}\right)\right\}\)

\(2\left(x-1\right)\left\{\left(x+\frac{3}{4}\right)^2-\left(2+\frac{9}{16}\right)\right\}=2\left(x-1\right)\left\{\left(x+\frac{3}{4}-2-\frac{9}{16}\right)\left(x+\frac{3}{4}+2+\frac{9}{16}\right)\right\}\)

\(=2x^3+4x^2-3x^2-6x+x+2\)

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

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

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

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

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

phan tích thành nhân tử nha

\(x^2+\frac{x}{2013}-\frac{2014}{2013}\)

\(x^2+\frac{2x}{4026}+\frac{1}{4026^2}-\left(\frac{2014}{2013}+\frac{1}{4026^2}\right)\)

\(\left(x+\frac{1}{4026}\right)^2-\left(\frac{2014}{2013}+\frac{1}{4026^2}\right)\)

\(\left(x+\frac{1}{4026}-\frac{2014}{2013}-\frac{1}{4026^2}\right)\left(x+\frac{1}{4026}+\frac{2014}{2013}+\frac{1}{4026^2}\right)\)