Question

[Lạc Chỉ]

a/ Nếu a + c = 2b

             2bd = c(b + d)

Thì a/b = c/d ( b, d \(\ne\)0)

b/ Nếu a² = bc

Thì a + b / a - b = c + a / c - a

Answer 1

[Laura]

Đề bài là cm à?

Ta có:

2bd=c(b+d)

=>(a+c)d=c(b+d)

=>ad+cd=cb+cd

=>ad+cd-cd=bc

=>ad=bc

=>a/b=c/d(đpcm)

Answer 2

[Xyz OLM]

a) Ta có \(\hept{\begin{cases}a+c=2b\left(1\right)\\2bd=c\left(b+d\right)\left(2\right)\end{cases}}\) 

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

\(\Rightarrow ad+cd=bc+cd\)

\(\Rightarrow ad=bc\)               

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

b) Ta có : a² = bc

=> \(\frac{a}{b}=\frac{a}{c}\)

Đặt \(\frac{a}{b}=\frac{a}{c}=k\)

=> a = bk = ck

Khi đó : \(\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)

\(\frac{c+a}{a-c}=\frac{c+ck}{ck-c}=\frac{c\left(1+k\right)}{c\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)

Từ (1) và (2) => \(\frac{a+b}{a-b}=\frac{c+a}{a-c}\left(\text{đpcm}\right)\)