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b^2 = a.c
=> a/b = b/c
Đặt a/b = b/c = k
=> a=bk ; b=ck
=> a = c.k.k = c.k^2 => a/c = k^2
Lại có : (a+2011b)^2/(b+2011c)^2
= (bk+2011b)^2/(ck+2011c)^2
= [b.(k+2011)]^2/[c.(k+2011)]^2
= b^2.(k+2011)^2/c^2.(k+2011)^2
= b^2/c^2
= (b/c)^2
= k^2
=> a/c = (a+2011)^2/(b+2011c)^2
Tk mk nha
\(b^2=ac\Rightarrow\frac{b}{c}=\frac{a}{b}=\frac{2010a}{2010b}=\frac{2011b}{2011c}=\frac{2010a+2011b}{2010b+2011c}\)
\(\Rightarrow\frac{b}{c}.\frac{a}{b}=\left(\frac{2010a+2011b}{2010b+2011c}\right).\left(\frac{2010a+2011b}{2010b+2011c}\right)\)
\(\Rightarrow\frac{a}{c}=\frac{\left(2010a+2011b\right)^2}{\left(2010b+2011c\right)^2}\)
Ta có: b2=ac\(\Rightarrow\frac{b}{c}=\frac{a}{b}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2016.b}{2016.c}\)(1)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{2016.b}{2016.c}=\frac{a+2016.b}{b+2016.c}\)(2)
Từ (1) và (2) ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{a+2016.b}{b+2016.c}\)
\(\Rightarrow\frac{\left(a+2016.b\right)^2}{\left(b+2016.c\right)^2}=\frac{a^2}{b^2}=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}\)(vì \(\frac{a}{b}=\frac{b}{c}\))\(=\frac{a}{c}\)(điều phải chứng minh)
Ta có VP:
\(\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
Thay \(1=ab+bc+ca\)
\(=\dfrac{2}{\sqrt{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}}\)
\(=\dfrac{2}{\sqrt{\left[b\left(a+c\right)+a\left(a+c\right)\right]\left[a\left(b+c\right)+b\left(b+c\right)\right]\left[b\left(a+c\right)+c\left(a+c\right)\right]}}\)
\(=\dfrac{2}{\sqrt{\left(a+c\right)\left(a+b\right)\left(a+b\right)\left(b+c\right)\left(b+c\right)\left(a+c\right)}}\)
\(=\dfrac{2}{\sqrt{\left[\left(a+c\right)\left(a+b\right)\left(b+c\right)\right]^2}}\)
\(=\dfrac{2}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)
_____________
Ta có VT:
\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\)
Thay \(1=ab+ac+bc\)
\(=\dfrac{a}{ab+ac+bc+a^2}+\dfrac{b}{ab+ac+bc+b^2}+\dfrac{c}{ab+ac+bc+c^2}\)
\(=\dfrac{a}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b}{b\left(b+c\right)+a\left(b+c\right)}+\dfrac{c}{c\left(b+c\right)+a\left(b+c\right)}\)
\(=\dfrac{a}{\left(a+c\right)\left(a+b\right)}+\dfrac{b}{\left(a+b\right)\left(b+c\right)}+\dfrac{c}{\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{a\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}+\dfrac{b\left(a+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\dfrac{c\left(a+b\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{ab+ac+ab+bc+ac+bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2ab+2ac+2bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2\cdot\left(ab+ac+bc\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\left(ab+ac+bc=1\right)\)
Mà: \(VP=VT=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\left(dpcm\right)\)
\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)
Giải:
Ta có: \(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=k\)
+) \(k^2=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}\) (1)
+) \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2011b}{2011c}=\dfrac{a+2011b}{b+2011c}\) ( t/c dãy tỉ số bằng nhau )
\(\Rightarrow k^2=\left(\dfrac{a+2011b}{b+2011c}\right)^2=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (2)
Từ (1), (2) \(\Rightarrow\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\left(đpcm\right)\)
Giải:
Từ hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b\) ta có:
\(VP=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}=\dfrac{a^2+2.2011ab+\left(2011b\right)^2}{b^2+2.2011bc+\left(2011c\right)^2}\)
\(=\dfrac{a^2+2.2011ab+2011^2ac}{ac+2.2011bc+2011^2c^2}\)
\(=\dfrac{a\left(a+2.2011b+2011^2c\right)}{c\left(a+2.2011b+2011^2c\right)}=\dfrac{a}{c}=VT\)
Vậy \(\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (Đpcm)