giúp mh nhanh vs

Ta có:

$\dfrac{1}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{abc+bc+b}$

$=\dfrac{abc}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$ (do $abc=1$)

$=\dfrac{abc}{a(bc+b+1)}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$

$=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$

$=\dfrac{bc+b+1}{bc+b+1}=1$

(đpcm)

Vì abc=1 nên:

\(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}=\frac{1}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{a}{abc.a+abc+ab}=\frac{1}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{a}{a+1+ab}=1\) 

Chúc bạn học tốt.

Lời giải:

Vì $a+b+c=1$ nên:

\(a^2+b^2+abc-1=(a+b)^2-2ab+abc-1\)

\(=(a+b)^2-1+ab(c-2)=(1-c)^2-1+ab(c-2)\)

\(=-c(2-c)+ab(c-2)=c(c-2)+ab(c-2)=(c+ab)(c-2)\)

Do đó:

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

Hoàn toàn tương tự với các phân thức còn lại, suy ra:

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

\(=\frac{ab+bc+ac-4(a+b+c)+12}{(a-2)(b-2)(c-2)}=\frac{ab+bc+ac+8}{(a-2)(b-2)(c-2)}\)

Ta có đpcm.

Akai Haruma

\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)

\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)

\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)

\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)