Đặt \(\left\{{}\begin{matrix}x^{671}=a\\y^{671}=b\end{matrix}\right.\). Bài toán trở thành

Cho \(a+b=0,67\) và \(a^2+b^2=1,34\). Tính \(A=a^3+b^3\)

Giải:

\(a^2+2ab+b^2=0,4489\)

\(\Rightarrow ab=\dfrac{0,4489-1,34}{2}=-0,44555\)

\(A=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=1,1963185\)

\(4B=\dfrac{4}{\sqrt{5}+1}+\dfrac{4}{\sqrt{6}+\sqrt{2}}+...+\dfrac{4}{\sqrt{2014}+\sqrt{2010}}\)

\(=\dfrac{4\left(\sqrt{5}-1\right)}{5-1}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}+...+\dfrac{4\left(\sqrt{2014}-\sqrt{2010}\right)}{2014-2010}\)

\(=\sqrt{5}-1+\sqrt{6}-\sqrt{2}+...+\sqrt{2014}-\sqrt{2010}\)

\(=-1-\sqrt{2}-\sqrt{3}-\sqrt{4}+\sqrt{2011}+\sqrt{2012}+\sqrt{2013}+\sqrt{2014}\)

\(\Rightarrow B=...\)

\(x^{671}+y^{671}=1\Rightarrow\left(x^{671}+y^{671}\right)^2=x^{1342}+2.x^{671}.y^{671}+y^{1342}\)\(=1\)

Mà \(x^{1342}+y^{1342}=2\) \(\Rightarrow x^{671}.y^{671}=\dfrac{-1}{2}\)

Mặt khác: \(\left(x^{671}+y^{671}\right)^3=x^{2013}+3x^{671}y^{671}\left(x^{671}+y^{671}\right)+y^{2013}=1\)

Hay \(x^{2013}+y^{2013}-\dfrac{3}{2}.1=1\Rightarrow x^{2013}+y^{2013}=1+\dfrac{3}{2}=\dfrac{5}{2}\)

Đặt \(\hept{\begin{cases}x^{671}=a\\y^{671}=b\end{cases}}\)thì ta có

\(\hept{\begin{cases}a+b=8,023\\a^2+b^2=32,801425\end{cases}}\)

\(\Rightarrow\left(a+b\right)^2=64,368529\)

\(\Leftrightarrow=ab=15,783552\)

Ta cần tính

\(F=\left(\frac{a^3+b^3}{2012}\right)^3-8,1234\)

\(=\left(\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{2012}\right)^3-8,1234\)

\(=\left(\frac{8,023.\left(32,801425-15,783552\right)}{2012}\right)^3-8,1234\)

\(=-8,12309\)                   

đề này dọa người thôi, máy tính mà ==" có thấy j khó =="

Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\) 

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

\(=\dfrac{1}{x+y+z}\)

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

giờ nhân cả tử và mẫu mỗi phân thức vs mỗi tử của nó rồi sử dụng BDT bunhiacopxki là ra thôi bn

\(\frac{x^2}{x^3-xyz+2013x}+\frac{y^2}{y^3-xyz+2013y}+\frac{z^2}{z^3-xyz+2013z}\)

\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+3.\left(xy+yz+zx\right)\left(x+y+z\right)}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx+3xy+3yz+3zx\right)}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x+y+z\right)^2}=\frac{1}{x+y+z}\)

\(VT=\text{Σ}_{cyc}\frac{x}{x^2-yz+2013}=\text{Σ}_{cyc}\frac{x^2}{x^3-xyz+2013x}\)

\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)(bđt Cauchy - Schwarz dạng Engel)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+2013\left(x+y+z\right)}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx+2013\right)}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+zx\right)+2013\right]}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[\left(x+y+z\right)^2-3.671+2013\right]}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)

(Dấu "=" xảy ra khi x = y = z = \(\frac{\sqrt{2013}}{3}\))