ab²+ac²+bc²+ba²+ca²+b²c+2abc

=bc²+b²c+ac²+abc+ab²+abc+ba²+ca²

=bc(b+c)+ac(b+c)+ab(b+c)+a²(b+c)

=(b+c)(bc+ac+ab+a²)

=(b+c)[c(a+b)+a.(a+b)]

=(b+c)(a+b)(a+c)

đặt y=x²-9 

Ta đc pt mới : (y+5)(y-5)=72

<=>y²-25=72<=>y²=97<=>y=căn 97 hoặc -căn 97

Thế y=x²-9 vào lại rồi tìm x

nghiệm có vẻ ko đẹp

M= (x+2)(x+4)(x+6)(x+8)+16

=(x+2)(x+8)(x+4)(x+6)+16

=(x²+10x+16)(x²+10x+24)+16

=(x²+10x+16)(x²+10x+16+8)+16

=(x²+10x+16)²+8(x²+10x+16)+16

=(x²+10x+20)²

=>dpcm

M=(x+2)(x+4)(x+6)(x+8)+16

=(x²+10x+16)(x²+10x+24)+16

=(x²+16+10x)(x²+10x+16+8)+16

=(x²+10x+16)²+8(x²+10x+16)+16

=(x²+10x+20)²

ĐPCM

bài dưới chỗ quy đồng thôi nha

\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)

<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x^3-1\right)\left(x^3+1\right)}\)

<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)}\)

<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

<=>\(\frac{\left(x^2-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}\)=\(\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

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

Quy đồng khử mẫu thôi mà :))

bài này c-s sẽ đỡ lo ngược hơn, nhưng trên có ghi am-gm thì xài am-gm thôi ( t cx hay bị ngược dấu vs lại dg muốn ngủ nên xét bài hộ)

Bài giải__

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{1-ab}=1+\frac{ab}{1-ab}\le1+\frac{ab}{1-\frac{a^2+b^2}{2}}=1+\frac{2ab}{a^2+b^2+2c^2}\)

\(=1+\frac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\le1+\frac{ab}{\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}}\)

\(\le1+\frac{1}{2}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)\). Tương tự ta cũng có:

\(\hept{\begin{cases}\frac{1}{1-bc}\le1+\frac{1}{2}\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)\\\frac{1}{1-ca}\le1+\frac{1}{2}\left(\frac{c^2}{b^2+c^2}+\frac{a^2}{a^2+b^2}\right)\end{cases}}\)

Cộng theo vế 3 BĐT ta dc: 

\(VT\le3+\frac{1}{2}\left(\frac{a^2+b^2}{a^2+b^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{c^2+a^2}{c^2+a^2}\right)=\frac{9}{2}\)

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

Làm ctv mà hong bít