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Ta có : \(\frac{2014a^2+b^2+c^2}{a^2}=\frac{a^2+2014b^2+c^2}{b^2}=\frac{a^2+b^2+2014c^2}{c^2}\)
\(\Rightarrow\) \(2014+\frac{b^2+c^2}{a^2}=2014+\frac{a^2+c^2}{b^2}=2014+\frac{a^2+b^2}{c^2}\)
\(\Rightarrow\) \(\frac{b^2+c^2}{a^2}=\frac{a^2+c^2}{b^2}=\frac{a^2+b^2}{c^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{b^2+c^2}{a^2}=\frac{a^2+c^2}{b^2}=\frac{a^2+b^2}{c^2}=\frac{2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=2\) (Vì \(a^2+b^2+c^2\ne0\))
Suy ra: \(\frac{b^2}{a^2}+\frac{c^2}{a^2}=\frac{a^2}{b^2}+\frac{c^2}{b^2}=\frac{a^2}{c^2}+\frac{b^2}{c^2}=2\)
\(\Rightarrow\frac{b^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{b^2}{c^2}=2+2+2=6\)
\(\Rightarrow\) \(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}=\frac{6}{2}=3\)
Lại có: \(P=\)\(\frac{2015a^2+b^2}{c^2}+\frac{2015a^2+c^2}{b^2}+\frac{2015b^2+c^2}{a^2}\)
\(=2015\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)+\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}\right)\)
\(=\left(2015+1\right)\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)\)
\(=2016\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)\)
\(=2016.3=6048\)
Vậy \(P=6048\)
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Ta có:
\(\dfrac{a}{c}=\dfrac{c}{b}\) \(\Rightarrow\) \(\dfrac{a^2}{c^2}=\dfrac{c^2}{b^2}\) \(=\dfrac{a^2+c^2}{c^2+b^2}\)
\(\dfrac{a}{b}=\dfrac{a}{c}\times\dfrac{c}{b}=\left(\dfrac{a}{c}\right)^2\)
Mà \(\dfrac{a^2+c^2}{b^2+c^2}=\left(\dfrac{a}{c}\right)^2=\dfrac{a}{b}\)
Vậy \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\) (ĐPCM)
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ta có
a/b=b/c
=>a.c=b^2
a^2+b^2/b^2+c^2
=a^2+a.c/a.c+c^2
=a(a+c)/c(a+c)
=a/c
k ung hộ mình nha
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Ta có: b2 = ac
a2 + b2/b2 + c2
= a2 + ac/ac + c2
= a.(a+c)/c.(a+c)
= a/c ( đpcm)
Theo đề ra ta có \(b^2=ac\)
\(a^2+\frac{b^2}{b^2}+c^2\)
\(=a^2+\frac{ac}{ac}+c^2\)
\(=a.\frac{\left(a+c\right)}{c}.\left(a+c\right)\)
Từ đó suy ra \(a^2+\frac{b^2}{b^2}+c^2=\frac{a}{c}\)
CHÚC BẠN HỌC TỐT !!!