vân hả bạn

Theo đề bài ta có:

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

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

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

\(\Rightarrow C=\dfrac{2017}{672}-3=\dfrac{1}{672}\)

a) Tính chất dãy tỉ số bằng nhau: \(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y+x-y}{2014+2016}=\dfrac{2x}{4030}=\dfrac{x}{2015}\)

\(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y-x+y}{2014-2016}=\dfrac{2y}{-2}=\dfrac{y}{-1}\)

Nên: \(\dfrac{x}{2015}=\dfrac{y}{-1}=\dfrac{xy}{2015}\)

Xét: \(\left\{{}\begin{matrix}\dfrac{x}{2015}=\dfrac{xy}{2015}\Leftrightarrow2015x=2015xy\Leftrightarrow y=1\\\dfrac{y}{-1}=\dfrac{xy}{2015}\Leftrightarrow2015y=-1xy\Leftrightarrow2015=-1x\Leftrightarrow x=-2015\end{matrix}\right.\)

2) \(VT=\left|x-6\right|+\left|x-10\right|+\left|x-2022\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT=\left|x-6\right|+\left|2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT\ge\left|x-6+2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(VT\ge2016+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\ge2016=VP\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}6\le x\le2022\\x=10\\y=2014\\z=2015\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=2014\\z=2015\end{matrix}\right.\)

a. (0,125)³ . 512

= \(\frac{1}{512}\).512

= 1

b. \(\left(\frac{90}{15}\right)^3\)

= 6³ = 216

c. [(0,1)³]

= \(\frac{1}{1000}\)

d. \(\left[\left(\frac{-1}{27}\right)^3\right]^6\)

= \(\left(-\frac{1}{27}\right)^{18}\)

\(\frac{a}{2}+\frac{b}{3}=\frac{3a+2b}{6}=\frac{a+b}{5}\)

\(\Rightarrow5\left(3a+2b\right)=6\left(a+b\right)\)

\(\Rightarrow15a+10b=6a+6b\)

\(\Rightarrow15a-6a=6b-10b\)

\(\Rightarrow9a=-4b\)\(\Rightarrow\frac{a}{-4}=\frac{b}{9}\)

Vì -4 < 0 ; 9 > 0 \(\Rightarrow\)a và b trái dấu

Vậy không tồn tại stn a, b 

a)Ta có :

\(\frac{4}{5}=\frac{8}{10}\)

1,1=\(\frac{11}{10}\)

Vì \(\frac{8}{10}< \frac{11}{10}\)

»\(\frac{4}{5}< 1,1\)

vậy \(\frac{4}{5}< 1,1\)

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