Chứng minh rằng:$\frac{a}{b}.\frac{a}{c}=\frac{a}{b}+\frac{a}{c}$ với $a;b;c$ là số nguyên, $b;c \neq 0$ và $b+c=a$
$\frac{a}{b}.\frac{a}{c}=\frac{a}{b}+\frac{a}{c}$
Started By Pham The Quang 6c, 10-03-2013 - 17:33
#2
Posted 17-03-2013 - 21:47
thay a=b+c vào biểu thức $\tfrac{a}{b}.\tfrac{a}{c}$ ta có $\tfrac{b+c}{b}.\tfrac{b+c}{c}=(1+\tfrac{c}{b})(1+\tfrac{b}{c})=1+\tfrac{c}{b}+\tfrac{b}{c}+\tfrac{bc}{bc}=2+\tfrac{b}{c}+\tfrac{c}{b}=(1+\tfrac{c}{b})+(1+\tfrac{b}{c})=\tfrac{b+c}{b}+\tfrac{b+c}{c}=\tfrac{a}{b}+\tfrac{a}{c}$
(vì b+c=a)
(vì b+c=a)
- vnmath98 likes this
#3
Posted 18-03-2013 - 12:46
$\frac{a}{b}.\frac{a}{c}-\frac{a}{b}=\frac{a}{b}(\frac{a}{c}-1)=\frac{a}{b}\frac{b}{c}=\frac{a}{c}$Chứng minh rằng:$\frac{a}{b}.\frac{a}{c}=\frac{a}{b}+\frac{a}{c}$ với $a;b;c$ là số nguyên, $b;c \neq 0$ và $b+c=a$
Ta có đpcm
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