Question

cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)

chứng minh:\(\frac{5a+5b}{5b}=\frac{c^2+cd}{cd}\)

Answer 1

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

\(\Rightarrow\frac{5a+5b}{5b}=\frac{5b\left(k+1\right)}{5b}=k+1\)

\(\frac{c^2+cd}{cd}=\frac{k^2d^2+kd^2}{kd^2}=\frac{kd^2\left(k+1\right)}{kd^2}=k+1\)

\(\Rightarrow\frac{5a+5b}{5b}=\frac{c^2+cd}{cd}\)
 

Answer 2

\(\)\(\frac{5a+5b}{5b}=\frac{5a}{5b}+\frac{5b}{5b}=\frac{a}{b}+1\)

\(\frac{c^2+cd}{cd}=\frac{c^2}{cd}+\frac{cd}{cd}=\frac{c}{d}+1\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{5a+5b}{5b}=\frac{c^2+cd}{cd}\)

\(\Rightarrowđpcm\)

Answer 3

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a= b.k ; c= d.k

\(\frac{5a+5b}{5b}=\frac{5a}{5b}+\frac{5b}{5b}=\frac{a}{b}+1=\frac{b.k}{b}+1=k+1\left(1\right)\)

\(\frac{c^2+cd}{cd}=\frac{\left(d.k\right)^2}{cd}+\frac{cd}{cd}=\frac{d^2.k^2}{d.k.d}+1=\frac{d.k}{d}+1=k+1\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrowđpcm\)

Answer 4

đặt a/b=c/d=k

=>a=bk

c=dk

5a+5b/5b=5bk+5b/5b=5b(k+1)/5b=k+1(1)

c^2+cd/cd=d^2.k^2+d^2.k/d^2.k=d^2.k(k+1)/d^2.k=k+1(2)

từ (1) và (2)=>đpcm