Ta có : \(a-b-c=0\)
\(\Rightarrow\left\{{}\begin{matrix}a=b+c\\b=a-c\\c=a-b\end{matrix}\right.\)
Thay a = b + c ; b = a - c ; c = a - b vào biểu thức A , ta được :
\(A=\left(1-\dfrac{a-b}{a}\right)\left(1-\dfrac{b+c}{b}\right)\left(1+\dfrac{a-c}{c}\right)\)
\(=\left(1-\dfrac{a}{a}+\dfrac{b}{a}\right)\left(1-\dfrac{b}{b}-\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}-\dfrac{c}{c}\right)\\ =\dfrac{b}{a}\cdot\dfrac{-c}{b}\cdot\dfrac{a}{c}=-1\)
Từ A= \(\left(1-\dfrac{c}{a}\right).\left(1-\dfrac{a}{b}\right).\left(1+\dfrac{b}{c}\right)\)ta có :a-b-c
Nên: a = b+c
b = a-c 1 \(\)
c = a-b
Thay 1 vào biểu thức A, ta có được:
A = \(A=\left(1-\dfrac{a-b}{a}\right).\left(1-\dfrac{b+c}{b}\right).\left(1+\dfrac{a-c}{c}\right)\) \(=\left(1-\dfrac{a}{a}+\dfrac{b}{a}\right).\left(1-\dfrac{b}{b}-\dfrac{c}{b}\right).\left(1+\dfrac{c}{c}-\dfrac{a}{c}\right)\)
\(=\dfrac{b}{a}.\dfrac{-c}{b}.\dfrac{a}{c}=-1\)
chắc vậy ai thấy đúng tick mik với nhé
