2020 x 2021 - 3031  = 2020 x ( 2 + 2019 ) - 3031 = 2020 x 2019 + 2020 x 2 - 3031 = 2019 x 2020 + 1009

Nên ( 2019 x 2020 + 1009 ) : ( 2020  x 2021 - 3031 ) = ( 2019 x 2020 + 1009 ):( 2019 x 2020 + 1009 )=1

a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)

= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)

= 1
@Nguyen Thi Ngoc Linh

Lời giải:
Đặt $\frac{x+y}{x-y}=a; \frac{y+z}{y-z}=b; \frac{z+x}{z-x}=c$

Bằng phép biến đổi tương đương cơ bản, ta chỉ ra được:

$ab+bc+ac=-1$

$\Leftrightarrow (a+b+c)^2-(a^2+b^2+c^2)=-2$

$\Leftrightarrow a^2+b^2+c^2=(a+b+c)^2+2\geq 2$

Ta sẽ đi chứng minh $a^{2020}+b^{2020}+c^{2020}>\frac{2^{1010}{3^{1009}}$
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Áp dụng BĐT AM-GM cho các số không âm:

\(\frac{a^{2020}}{a^{2020}+b^{2020}+c^{2020}}+\frac{1}{3}+\frac{1}{3}+....+\frac{1}{3}\geq 1010\sqrt[1010]{\frac{a^{2020}}{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}}\)

\(\frac{b^{2020}}{a^{2020}+b^{2020}+c^{2020}}+\frac{1}{3}+\frac{1}{3}+....+\frac{1}{3}\geq 1010\sqrt[1010]{\frac{b^{2020}}{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}}\)

\(\frac{c^{2020}}{a^{2020}+b^{2020}+c^{2020}}+\frac{1}{3}+\frac{1}{3}+....+\frac{1}{3}\geq 1010\sqrt[1010]{\frac{c^{2020}}{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}}\)

Cộng theo vế và thu gọn: $a^2+b^2+c^2\leq \sqrt[1010]{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}$

$\Rightarrow a^{2020}+b^{2020}+c^{2020}\geq \frac{(a^2+b^2+c^2)^{1010}}{3^{1009}}\geq \frac{2^{1010}}{3^{1009}}$ do $a^2+b^2+c^2\geq 2$

Dấu "=" xảy ra khi $a=b=c$ và $a^2+b^2+c^2=2$. Điều này không được vì $x,y,z$ đôi một khác nhau làm $a,b,c$ đôi một khác nhau

Ta có đpcm.

Akai Haruma dạ giúp em bài này vs ạ !!!

\(\dfrac{x-2}{2016}+\dfrac{x-4}{1009}+\dfrac{x-6}{2020}=-4\)

\(\Leftrightarrow\) \(\dfrac{x-2}{2016}+1+\dfrac{x-4}{1009}+2+\dfrac{x-6}{2020}+1=0\)

\(\Leftrightarrow\) \(\dfrac{x-2+2016}{2016}+\dfrac{x-4+2018}{1009}+\dfrac{x-6+2020}{2020}=0\)

\(\Leftrightarrow\) \(\dfrac{x-2014}{2016}+\dfrac{x-2014}{1009}+\dfrac{x-2014}{2020}=0\)

\(\Leftrightarrow\) \(\left(x-2014\right)\left(\dfrac{1}{2016}+\dfrac{1}{1009}+\dfrac{1}{2020}\right)=0\)

\(\Leftrightarrow\) x - 2014 = 0

\(\Leftrightarrow\) x = 2014

Vậy............

1/\(-2020+23+x=-2020\\ \Leftrightarrow23+x=-2020+2020\\ \Leftrightarrow23+x=0\\ \Leftrightarrow x=0-23\\ \Leftrightarrow x=-23\)

Vậy...

2/\(2x-35=25\\ \Leftrightarrow2x=60\\ \Leftrightarrow x=30\)

Vậy...

3/\(3x+17=26\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\)

Vây...

4/\(\left|\text{x}-1\right|=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)

Vậy...

5/ \(-17.\left|x\right|=-34\\ \Leftrightarrow\left|x\right|=2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

VẬy...

\(\dfrac{x-2}{2016}+\dfrac{x-4}{1009}+\dfrac{x-6}{2020}=-4\)

<=>\(\dfrac{x-2}{2016}+1+\dfrac{x-4}{1009}+2+\dfrac{x-6}{2020}+1=0\)

<=>\(\dfrac{x+2014}{2016}+\dfrac{x+2014}{1009}+\dfrac{x+2014}{2020}=0\)

<=>\(\left(x+2014\right)\left(\dfrac{1}{2016}+\dfrac{1}{1009}+\dfrac{1}{2020}\right)=0\)

vì 1/2016+1/1009+1/2020 khác 0

=>x+2014=0<=>x=-2014