ta có a/b= 4.5/7 bằng với a/4.5=b/7

a/4.5=b/7=a-b/4.5-7=-5/-2.5=2

a=4.5x2=9

b=7x2=14

vậy a=9 b=14

ngu vai bai de the cung ko biet lam oc bo

bà ngu day làm sao, éo tích cho luôn bây giờ:<

ta có :

\(A=\frac{a^2}{1-a}+a+\frac{b^2}{1-b}+b+\frac{1}{a+b}=\frac{a}{1-a}+\frac{b}{1-b}+\frac{1}{a+b}\)

\(A=\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{a+b}-2\)

mà : \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{a+b}\ge\frac{9}{1-a+1-b+a+b}=\frac{9}{2}\)

Vậy \(A\ge\frac{9}{2}-2=\frac{5}{2}\)

dấu bằng xảy ra khi : \(1-a=1-b=a+b\Leftrightarrow a=b=\frac{1}{3}\)

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

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

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

Đặt \(\left(\frac{a}{b^2},\frac{b}{c^2},\frac{c}{a^2}\right)=\left(x,y,z\right)\)

\(\Rightarrow xyz=\frac{abc}{a^2b^2c^2}=\frac{1}{abc}=1\)

Theo bài ra ta có : \(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\)

\(\Leftrightarrow x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow x+y+z=xy+yz+xz\)

\(\Leftrightarrow\left(xy-x-y+1\right)-1+z\left(x+y-1\right)=0\)

\(\Leftrightarrow\left(xy-x-y+1\right)+z\left(x+y-1-xy\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)-z\left(x-1\right)\left(y-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(1-z\right)=0\)

\(\Leftrightarrow\frac{a-b^2}{b^2}.\frac{b-c^2}{c^2}.\frac{a^2-c}{a^2}=0\)

\(\Leftrightarrow\left(a-b^2\right)\left(b-c^2\right)\left(c-a^2\right)=0\)

Ta có đpcm

mình ko biết

Ta có:

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{b}+\frac{1}{d}\right)=\frac{b+d}{2bd}\)

\(\Rightarrow2bd=c\left(b+d\right)\left(2\right)\)

Do b là TBC của a và c nên \(b=\frac{a+c}{2}\)

Thay vào (1) ta có: \(2.\frac{a+c}{2}.d=c.\left(\frac{a+c}{2}+d\right)\)

=> (a + c).d = \(\frac{c.\left(a+c+2d\right)}{2}\)

=> (a + c).2d = c.(a + c + 2d)

=> 2ad + 2cd = ac + c² + 2cd

=> 2ad = ac + c² = c.(a + c) = c.2b

=> ad = bc

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

8,8