Question

cho 3 số a,b,c khác 0 và a-b-c=0 tính giá trị biểu thức Q=(1-c/a)*(1-a/b)*(1+b/c)

Answer 1

\(\left(1-\dfrac{c}{a}\right)\left(1-\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\)

\(=\left(\dfrac{a-c}{a}\right)\left(\dfrac{b-a}{b}\right)\left(\dfrac{b+c}{c}\right)\) \(\left(1\right)\)

Vì \(a-b-c=0\Rightarrow\left\{{}\begin{matrix}a-c=b\\b-a=-c\\b+c=a\end{matrix}\right.\) \(\left(2\right)\)

Thay (2) vào (1) ta có:

\(\left(\dfrac{a-c}{a}\right)\left(\dfrac{b-a}{b}\right)\left(\dfrac{b+c}{c}\right)\)

\(=\dfrac{b}{a}.\dfrac{\left(-c\right)}{b}.\dfrac{a}{c}=-1\)

Vậy S=-1

Answer 2

ối giàng ơi

Answer 3

haha