\(K=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017bc}{abc+2017bc+2017b}\)

\(K=\frac{a}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{2017bc}{2017\left(bc+b+1\right)}\)

\(K=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=\frac{bc+b+1}{bc+b+1}=1\)

\(H=\frac{4x^2+4-4x^2-4x-1}{x^2+1}=\frac{4\left(x^2+1\right)}{x^2+1}-\frac{\left(2x+1\right)^2}{x^2+1}=4-\frac{\left(2x+1\right)^2}{x^2+1}\le4\)

\(\Rightarrow H_{max}=4\) khi \(x=-\frac{1}{2}\)

a) \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)

\(=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{4x^2-3x+17+2x^2-x-2x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=-\frac{12}{x^2+x+1}\)

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

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

c) \(N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)

\(N=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)

\(N=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)

\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac+abc^2+abc}\)

\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac\left(1+bc+b\right)}\)

\(N=\frac{1+b}{b+1+bc}+\frac{bc}{1+bc+b}\)

\(N=\frac{1+b+bc}{b+1+bc}\)

\(N=1.\)

Ta có:

\(\frac{ab}{\sqrt{2017c+ab}}=\frac{ab}{\sqrt{\left(a+b+c\right)c+ab}}\)

\(=\frac{ab}{\sqrt{a\left(b+c\right)+c\left(b+c\right)}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

Áp dụng BĐT AM-GM (cô si): \(ab.\frac{1}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{ab}{2}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{ab}{2\left(a+c\right)}+\frac{ab}{2\left(b+c\right)}\)

Tương tự với hai BĐT còn lại và cộng theo vế,ta được:

\(A\le\frac{ab}{2\left(a+c\right)}+\frac{ab}{2\left(b+c\right)}+\frac{bc}{2\left(a+b\right)}+\frac{bc}{2\left(a+c\right)}+\frac{ca}{2\left(b+c\right)}+\frac{ca}{2\left(a+b\right)}\)

Thu gọn lại bằng cách cộng những phân thức cùng mẫu và rút gọn phân thức,ta được:

\(A\le\frac{a+b+c}{2}=\frac{2017}{2}\).

Dấu "=" xảy ra khi \(a=b=c=\frac{2017}{3}\)

Vậy...

Có vài cách giải nhưng mình thấy cách này nhanh và đẹp ne.

\(\sqrt{2017a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(c+a\right)}\le\sqrt{ac}+\sqrt{ab}\)

\(\Rightarrow\frac{a}{a+\sqrt{2017a+bc}}\le\frac{a}{a+\sqrt{ab}+\sqrt{bc}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự rồi cộng lại, ta được:

\(P\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

Dấu "=" khi \(a=b=c=\frac{2017}{3}\)

a. Từ tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

Ta có: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\times\frac{b}{d}=\left(\frac{a-c}{b-d}\right)\left(\frac{a-c}{b-d}\right)=\left(\frac{a-c}{b-d}\right)^2\)

\(\Rightarrow\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)(ĐPCM)

a)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\) Đặt \(\frac{a}{c}=\frac{b}{d}=k\)

Áp dụng TCDSBN ta có :

\(k=\frac{a-b}{c-d}\)\(\Rightarrow k^2=\left(\frac{a-b}{c-d}\right)^2\)(1)

Ta lại có : \(k=\frac{a}{c};k=\frac{b}{d}\Rightarrow k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)(2)

Từ (1) ; (2) \(\Rightarrow\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\)(đpcm)

b ) Đề sai : điều cần cm là \(\frac{2017a-2018b}{2017c+2018d}=\frac{2017c-2018d}{2017a+2018b}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2007a}{2007c}=\frac{2008b}{2008c}=\frac{2007a+2008b}{2007c+2008d}=\frac{2007a-2008b}{2007c-2008d}\)

\(\Rightarrow\left(2007a+2008b\right)\left(2007c-200d\right)=\left(2007a-2008b\right)\left(2007c+2008d\right)\)

\(\Rightarrow\frac{2017a-2018b}{2017c+2018d}=\frac{2017c-2018d}{2017a+2018b}\)(đpcm)

\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}\)

Áp dụng tính chất dãy tỉ số bằng nhau: 

\(\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2017a-2018b}{2017c-2018d}=\frac{2018a+2019b}{2018c+2019d}\)

<=>\(\left(2017a-2018b\right)\left(2018c+2019d\right)=\left(2018a+2019b\right)\left(2017c-2018d\right)\)

<=>\(\frac{2017a-2018b}{2018a+2019b}=\frac{2017c-2017d}{2018x+2019d}\)(đpcm)

nhật gà