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Lời giải:
Có: \(\left\{\begin{matrix} a+b+c=9\\ a^2+b^2+c^2=27\end{matrix}\right.\Rightarrow \left\{\begin{matrix} (a+b+c)^2=81\\ a^2+b^2+c^2=27\end{matrix}\right.\)
\(\Rightarrow (a+b+c)^2-(a^2+b^2+c^2)=54\)
\(\Leftrightarrow 2(ab+bc+ac)=54\Leftrightarrow ab+bc+ac=27\)
Do đó: \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow \frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}=0(*)\)
Ta thấy: \((a-b)^2; (b-c)^2; (c-a)^2\geq 0\forall a,b,c\in\mathbb{R}\)
Suy ra \((*)\) xảy ra khi và chỉ khi
\((a-b)^2=(b-c)^2=(c-a)^2=0\Leftrightarrow a=b=c\)
Khi đó: \(a=b=c=\frac{9}{3}=3\) (thỏa mãn)
\(P=(a-2)^{2015}+(b-3)^{2016}+(c-4)^{2017}=1^{2015}+0^{2016}+(-1)^{2017}\)
\(P=1+0+(-1)=0\)
\(P=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\left(\sqrt{a}+4\right)}\)
\(=\dfrac{-8\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}\cdot\dfrac{\sqrt{a}+4}{-\left(\sqrt{a}+1\right)}=\dfrac{8}{\sqrt{a}-4}\)
P= (\(\frac{3\sqrt{a}}{\sqrt{a}+4}+\frac{\sqrt{a}}{\sqrt{a}-4}+\frac{4\left(a+2\right)}{16-a}\)):\(\left(1-\frac{2\sqrt{a}+5}{\sqrt{a}-4}\right)\)
=\(\left(\frac{3\sqrt{a}\left(\sqrt{a}-4\right)}{a-16}+\frac{\sqrt{a}\left(\sqrt{a}+4\right)}{a-16}-\frac{4a+8}{a-16}\right):\left(\frac{\sqrt{a}-4-2\sqrt{a}-5}{\sqrt{a}-4}\right)\)
= \(\left(\frac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{a-16}\right):\left(\frac{-\sqrt{a}-9}{\sqrt{a}-4}\right)\)
=\(\left(\frac{-8\sqrt{a}-8}{a-16}\right).\left(\frac{\sqrt{a}-4}{-\sqrt{a}-9}\right)=\frac{8\sqrt{a}+8}{\left(\sqrt{a}+4\right).\left(\sqrt{a}+9\right)}=\frac{8\sqrt{a}+8}{a+13\sqrt{a}+36}\)
Với mọi n>0 ta có:\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng đẳng thức trên vào D ta được:
\(D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}=1-\frac{\sqrt{2016}}{2016}=\frac{2016-\sqrt{2016}}{2016}\)
Ta có:
\(\frac{5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2}{\left(y-x\right)^2}=\frac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(x-y\right)^2}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4=5\left(x^2+2xy+y^2\right)-3x+3y+4\)
\(=5x^2+10xy+5y^2-3x+3y+4\)
Ta có: \(B=4^{2017}+4^{2016}+...+4^2+4^1+4^0\)
\(\Leftrightarrow4\cdot B=4^{2018}+4^{2017}+...+4^3+4^2+4^1\)
\(\Leftrightarrow3\cdot B=4^{2018}-1\)
\(\Leftrightarrow A=165\cdot\dfrac{4^{2018}-1}{3}+55\)
\(\Leftrightarrow A=4^{2018}\)