Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)

Ta có: \(\left(\frac{a-b}{c-d}\right)^4=\left(\frac{bk-b}{dk-d}\right)^4=\left[\frac{b\left(k-1\right)}{d\left(k-1\right)}\right]^4=\left(\frac{b}{d}\right)^4\) (1)

\(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\frac{b^4.k^4+b^4}{d^4.k^4+d^4}=\frac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\frac{b^4}{d^4}=\left(\frac{b}{d}\right)^4\) (2)

Từ (1) và (2) \(\Rightarrow\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\left(đpcm\right)\)

Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{a^4}{c^4}=\frac{b^4}{d^4}=\left(\frac{a-b}{c-d}\right)^4\) (1)

Ta lại có:

\(\frac{a^4}{c^4}=\frac{b^4}{d^4}=\frac{a^4+b^4}{c^4+d^4}\) (2)

Từ (1);(2)\(\Rightarrow\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)

Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)

\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)

ta sẽ giết ngươi kí tên dép đờ kiu lờ

Mình chỉ làm bài 1a, và bài 3 thôi nhé,còn lại là bạn tự làm nhé

Bài 1:

a, Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

\(\Rightarrow\left[\frac{a}{b}\right]^2=\left[\frac{c}{d}\right]^2=\left[\frac{a+c}{b+d}\right]^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{(a+c)^2}{(b+d)^2}\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}\)

Bài 3 : Sửa đề : Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

CM : a = b = c

Cách 1 : Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

vì \(a+b+c\ne0\)

\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c\)

Do đó : \(a=b=c\).

Cách 2 : Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=m\), ta có : \(a=bm,b=cm,c=am\)

Do đó : \(a=bm=m(mc)=m\left[m(ma)\right]\)

\(\Rightarrow a=m^3a\Rightarrow m^3=1(a\ne0)\Rightarrow m=1\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

Cách 3 : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}=\left[\frac{a}{b}\right]^3\Rightarrow1=\left[\frac{a}{b}\right]^3\Rightarrow\frac{a}{b}=1\)

Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)

Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)

Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)

\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)

Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*

\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{​​}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)

\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)

Đẳng thức xảy ra khi a = b = c

P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:

1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)

\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)

bài 2 xem có ghi nhầm ko

3a biến đổi tí là xong

b tuong tự bài 1 

a, \(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\Rightarrow\frac{a^4}{b^4}=\frac{c^4}{d^4}=\frac{\left(a-c\right)^4}{\left(b-d\right)^4}\) (1)

\(\frac{a^4}{b^4}=\frac{c^4}{d^4}=\frac{5a^4}{5b^4}=\frac{7c^4}{7d^4}=\frac{5a^4+7c^4}{5b^4+7d^4}\)(2)

Từ (1) và (2) => đpcm

b, \(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+2c}{b+2d}\) (3)

\(\frac{a}{b}=\frac{c}{d}=\frac{3c}{3d}=\frac{a-3c}{b-3d}\) (4)

Từ (3) và (4) => đpcm

c, làm giống câu a

a) ta có \(\frac{a}{b}=\frac{c}{d}=\frac{a+2c}{b+2d}\left(1\right)\)

            \(\frac{a}{b}=\frac{c}{d}=\frac{a-3c}{b-3d}\left(2\right)\)

(1) và (2) => \(\frac{a+2c}{b+2d}=\frac{a-3c}{b-3d}\)

Tương tự \(\left(\frac{a}{b}\right)^4=\left(\frac{c}{d}\right)^4=\left(\frac{a-c}{b-d}\right)^4\left(1\right)\)

               \(\left(\frac{a}{b}\right)^4=\left(\frac{c}{d}\right)^4=\frac{5a^4+7c^4}{5b^4+7d^4}\left(2\right)\)

=> \(\left(\frac{a-c}{b-d}\right)^4=\frac{5a^4+7c^4}{5b^4+7d^4}\)