I DON NO

a/b = c/d

=> a = bk và c = dk

thay vào ta có : 

(bk + 2dk)(b+d) = (bk+dk)(b+2d)

=> k(b+2d)(b+d) = k(b+d)(b+2d)

xong rồi nha

K bt thì k cần phải nói 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a = bk ; c = dk

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

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

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

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

AP DUNG T/C ...................................

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

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

\(\frac{\left(1\right),\left(2\right)}{\Rightarrow}\)dieu can chung minh

1.

\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)

\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)

Dấu "=" khi \(a=b=c\)

2.

\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)

Dấu "=" khi \(a=b=c=d\)

Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,

Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana

Help me!

Bài thứ hai đó áp dụng bđt cauchy showas là ra rồi sử dụng tch bắc cầu tệ.

ta có: \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\frac{a}{b}=k\Rightarrow a=bk\)

\(\frac{c}{d}=k\Rightarrow c=dk\)

thay vào \(\left(a+2c\right).\left(b+d\right)=\left(bk+2dk\right).\left(b+d\right)=k.\left(b+2d\right).\left(b+d\right)\)

\(\left(a+c\right).\left(b+2d\right)=\left(bk+dk\right).\left(b+2d\right)=k.\left(b+d\right).\left(b+2d\right)\)

\(\Rightarrow\left(a+2c\right).\left(b+d\right)=\left(a+c\right).\left(b+2d\right)\left(=k.\left(b+2d\right).\left(b+d\right)\right)\)( đ p c m)

CHÚC BN HỌC TỐT!!!!!!!!

Ta có:

\(\left(a+2c\right).\left(b+d\right)=\left(a+c\right).\left(b+2d\right)\)

\(ab+ad+2cb+2cd=ab+2ad+cb+2cd\)

\(cb=ad\Rightarrow\frac{a}{b}=\frac{c}{d}\)

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

Áp dụng tính chất của dãy tỉ số bằng nhau ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

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

b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow1+\frac{a}{b}=1+\frac{c}{d}\Rightarrow\frac{b+a}{b}=\frac{d+c}{d}\)

vậy \(\frac{a+b}{b}=\frac{c+d}{d}\)

a) Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\Rightarrow1-\frac{b}{a}=1-\frac{d}{c}\Rightarrow\frac{a-b}{a}=\frac{c-d}{c}\)

b) Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\Rightarrow1+\frac{b}{a}=1+\frac{d}{c}\Rightarrow\frac{a+b}{a}=\frac{c+d}{c}\)