Chứng minh rằng nếu \(c^2+2.\left(ab-ac-bc\right)=0\)và \(b\ne c\), \(a+b\ne c\)thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
7 tháng 5 2018
Do \(c^2+2\left(ab-ac-bc\right)=0\Leftrightarrow-c^2=2\left(ab-ac-bc\right)\)
Ta có; \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+c^2-c^2+\left(a-c\right)^2}{b^2+c^2-c^2+\left(b-c\right)^2}=\frac{a^2+c^2+2\left(ab-ac-bc\right)+\left(a-c\right)^2}{b^2+c^2+2\left(ab-ac-bc\right)+\left(b-c\right)^2}\)
\(=\frac{2\left(a-c\right)^2+2\left(ab-bc\right)}{2\left(b-c\right)^2+2\left(ab-ac\right)}=\frac{2\left(a-c\right)^2+2b\left(a-c\right)}{2\left(b-c\right)^2+2a\left(b-c\right)}=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}\)
\(=\frac{a-c}{b-c}\) (đpcm)
5 tháng 11 2016
Vì \(c^2+2\left(ab-ac-bc\right)=0\) nên :
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+\left(a-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}{b^2+\left(b-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}\)
\(=\frac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}=\frac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\) \(\left(b\ne c,a+b\ne0\right)\)
15 tháng 11 2017
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+\left(a-c\right)^2+c^2+2\left(ab-ac-bc\right)}{b^2+\left(b-c\right)^2+c^2+2\left(ab-ac-bc\right)}\)
\(=\dfrac{a^2+a^2-2ac+c^2+c^2+2ab-2ac-2bc}{b^2+b^2-2bc+c^2+c^2+2ab-2ac-2bc}\)
\(=\dfrac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}\)
\(=\dfrac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\dfrac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(a-c+b\right)}=\dfrac{a-c}{b-c}\left(đpcm\right)\)
13 tháng 11 2018
Với điều kiện như đề bài
Ta có: \(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{b^2-a^2+a^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{\left(b-a\right)\left(b+a\right)+\left(a-c\right)\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}=\frac{b-a}{a+c}+\frac{a-c}{a+b}\)
Tướng tự:
\(\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}=\frac{c-b}{b+a}+\frac{b-a}{b+c}\)
\(\frac{a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{a-c}{c+b}+\frac{c-b}{c+a}\)
Em nhớ làm tiếp nhé!
