cho b^2=ac . CMR a^2 + b^2 / b^2 + c^2 =a/c
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(VT=\dfrac{a^2+b^2}{b^2+c^2}\)
Mà \(b^2=ac\)
\(\Leftrightarrow VT=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}=VP\left(đpcm\right)\)
Vậy...
Ta có: \(b^2=ac\)
\(\Leftrightarrow ac=b\cdot b\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Leftrightarrow\dfrac{a^2}{ac}=\dfrac{a^2+b^2}{b^2+c^2}\)
hay \(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Leftrightarrow a=b=c=1\)
b) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ac\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2+b^2-2ab\right)+\left(b^2+c^2-2bc\right)+\left(c^2+a^2-2ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)