\(\left(x+y+z\right).\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{2017}{672}\)

\(\Rightarrow\left(\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}\right)=\dfrac{2017}{672}\)

\(\Rightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{x+z}=\dfrac{2017}{672}\)

\(\Rightarrow3+\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}\)

\(\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}-3=\dfrac{2017}{672}-\dfrac{2016}{672}=\dfrac{1}{672}\)

\(\Rightarrow C=\dfrac{1}{672}\)

ĐÁP ÁN: B

\(VT=\sqrt{\left(\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\right)^2}\)

\(\le\sqrt{3\left(x+y+z+3\right)}=\sqrt{\left[9-2\left(x+y+z\right)\right]+5\left(x+y+z\right)}\)

\(=\sqrt{\left[9-\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right]+5\left(x+y+z\right)}\le\sqrt{5\left(x+y+z\right)}=VP\)

Đẳng thức xảy ra khi \(a=b=c=\frac{3}{2}\)

Theo giả thiết \(2=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\Rightarrow x+y+z\ge\frac{9}{2}\)

\(\Rightarrow\frac{2}{3}\left(x+y+z\right)\ge3\)

\(VT=\sqrt{\left(\Sigma_{cyc}\frac{\sqrt{x+1}}{\sqrt{5\left(x+y+z\right)}}.\sqrt{5\left(x+y+z\right)}\right)^2}\le\sqrt{15\left(x+y+z\right)\left[\Sigma_{cyc}\frac{x+1}{5\left(x+y+z\right)}\right]}\)

\(=\sqrt{3\left(x+y+z+3\right)}\le\sqrt{3\left(x+y+z+\frac{2}{3}\left(x+y+z\right)\right)}=\sqrt{5\left(x+y+z\right)}=VP\)

Ta có: \(\dfrac{1992x}{xy+1992x+1992}\)=

\(\dfrac{xyz.x}{xy+xyz.x+xyz}\) = \(\dfrac{xyz.x.z}{xy.z+xyz.x.z+xyz.z}\) = \(\dfrac{xz}{1+xz+z}\)

Ta có: \(\dfrac{y}{zy+y+1992}\)=\(\dfrac{y}{zy+y+xyz}\)=\(\dfrac{1}{z+1+xz}\)

=> \(\dfrac{1992x}{xy+1992x+1992}\)+\(\dfrac{y}{zy+y+1992}\)+\(\dfrac{z}{z+zx+1}\) = \(\dfrac{xz}{1+zx+z}\) +\(\dfrac{1}{z+zx+1}\) \(+\dfrac{z}{z+zx+1}\) =\(\dfrac{z+zx+1}{z+xz+1}\)

=1

Đáp án A

z = − 1 + i − i 2 + i 3 − i 4 + i 5 − ... + i 99 − i 100 + i 101 = − 1 + i + i 2 − 1 + i + i 4 − 1 + i + .... + i 100 − 1 + i = − 1 + i 1 + i 2 + i 4 + ... + i 100 = − 1 + i 1 1 − i 2.51 1 − i 2 = − 1 + i .

Theo Viet: \(\left\{{}\begin{matrix}z_1+z_2=2i\\z_1z_2=-1-2i\end{matrix}\right.\)

\(\Rightarrow z_1^3+z_2^3=\left(z_1+z_2\right)\left(z_1^2+z_2^2-z_1z_2\right)=\left(z_1+z_2\right)\left(\left(z_1+z_2\right)^2-3z_1z_1\right)\)

\(=2i\left[\left(2i\right)^2-3\left(-1-2i\right)\right]=2i\left(6i-1\right)=-12-2i\)

Lời giải:

Áp dụng BĐT Cô - si:

\(P=ax^m+\frac{b}{x^n}=\frac{a}{n}x^m+\frac{a}{n}x^m+...+\frac{a}{n}x^m+\frac{b}{mx^n}+...+\frac{b}{mx^n}\)

\(=(m+n)\sqrt[m+n]{(\frac{a}{n})^n.x^{mn}.(\frac{b}{m})^m.\frac{1}{x^{mn}}}\)

\(=(m+n)\sqrt[m+n]{\frac{a^nb^m}{n^n.m^m}}\)

Ta co:

\(P=\Sigma_{cyc}\frac{x}{x+1}=3-\Sigma_{cyc}\frac{1}{x+1}\le3-\frac{9}{x+y+z+3}=\frac{9}{4}\)

Dau '=' xay ra khi \(x=y=z=\frac{1}{3}\)

Đáp án D.

Ta có:

log x + y = z log x 2 + y 2 = z + 1 ⇔ x + y = 10 z + x 2 + y 2 = 10 z + 1 = 10.10 z ⇒ x 2 + y 2 = 10 x + y

Khi đó:

x 3 + y 3 = a .10 3 z + b .10 2 z ⇔ x + y x 2 − x y + y 2 = a . 10 z 3 + b . 10 z 2 ⇔ x + y x 2 − x y + y 2 = a . x + y 3 + b . x + y 2 ⇔ x 2 − x y + y 2 = a . x + y 2 + b . x + y ⇔ x 2 − x y + y 2 = a . x 2 + 2 x y + y 2 + b 10 . x 2 + y 2 ⇔ x 2 + y 2 − x y = a + b 10 . x 2 + y 2 + 2 a . x y

Đồng nhất hệ số, ta được:

a + b 10 = 1 2 a = − 1 ⇒ a = − 1 2 b = 15 .

Vậy  a + b = 29 2 .