$(x+2019)\; x4037=(x+2018)\; x4038$

$\Rightarrow 4037x+\left(4037\times 2019\right)=4038x+\left(4038\times 2018\right)$

$\Rightarrow 4037x+8150703=4038x+8148684$

$\Rightarrow 4037x-4038x=-8150703+8148684$

⇒−�=−2019

$\Rightarrow x=2019$

$x+2019)\; x4037=(x+2018)\; x4038$

$\Rightarrow 4037x+\left(4037\times 2019\right)=4038x+\left(4038\times 2018\right)$

$\Rightarrow 4037x+8150703=4038x+8148684$

$\Rightarrow 4037x-4038x=-8150703+8148684$

⇒−�=−2019

$\Rightarrow x=2019$

$1\; tick\; với\; nha$

\(\frac{x+2019}{x+2018}=\frac{4038}{4037}\)

\(\Rightarrow\left(x+2019\right)4037=\left(x+2018\right)4038\)

\(\Rightarrow4037x+\left(4037\times2019\right)=4038x+\left(4038\times2018\right)\)

\(\Rightarrow4037x+8150703=4038x+8148684\)

\(\Rightarrow4037x-4038x=-8150703+8148684\)

\(\Rightarrow-x=-2019\)

\(\Rightarrow x=2019\)

P/s: Số to kinh -_- Ko chắc đúng đâu.

x = 4038 - 2019 = 2019

~Học tốt~ 

#ngocyen#

\(\frac{x+2019}{x+2018}=\frac{4038}{4037}\)

\(\Leftrightarrow4037(x+2019)=4038(x+2018)\)

\(\Leftrightarrow4037x+8150703=4038x+8148684\)

\(\Leftrightarrow4037x+8150703-4038x=8148684\)

\(\Leftrightarrow4037x-4038x+8150703=8148684\)

\(\Leftrightarrow-x=-2019\)

\(\Leftrightarrow x=2019\)

\(\frac{x+2019}{x+2018}=\frac{4038}{4037}\)

\(\Rightarrow4037.\left(x+2019\right)=4038.\left(x+2018\right)\)

\(4037x+8150703=4038x+8148684\)

\(4038x-4037x=8150703-8148684\)

\(x=2019\)

Hc tốt

/ là chia hay phần

phần đó bạn

\(x+y+z=2018\)\(\Rightarrow\)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2018}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\\ \Leftrightarrow x^2y+xy^2+xyz+xyz+y^2z+\\ yz^2+zx^2+xyz+z^2x-xyz=0\)

\(\Leftrightarrow x^2y+xy^2+xyz+xyz+\\ y^2z+yz^2+zx^2+z^2x=0\)

\(\Leftrightarrow xy\left(x+y\right)+yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)\left(xy+yz+xz+z^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y\left(x+z\right)+z\left(x+z\right)\right)=0\\ \Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

suy ra x+y=0 hoặc y+z=0 hoặc x+z=0

hay x=-y hoặc y=-z hoặc x=-z

thay vào D ta tính dc kq

Ta chứng minh 1 bổ đề sau: Với a;b lớn hơn hoặc bằng 1 thì \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

Thật vậy: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{a^2+b^2+2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(a^2+b^2+2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow a^2+a^3b+b^2+b^3a+2+2ab\ge2a^2+2b^2+2a^2b^2+2\)

\(\Leftrightarrow a^3b+b^3a+2ab-a^2-b^2-2a^2b^2\ge0\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)(đúng với a;b>=1)

Trở lại bđt trong bài: \(\frac{2019}{2019+x^2}+\frac{2019}{2019+y^2}\ge\frac{4038}{2019+xy}\)

\(\Leftrightarrow\frac{1}{2019+x^2}+\frac{1}{2019+y^2}\ge\frac{2}{2019+xy}\) bđt này tương tự với bđt vừa cm trong bài,với x;y là hoán vị của a;b và 2019 có vai trò như 1

a)

208 − 8. ( x + 4 ) = 2 3 .5 8. ( x + 4 ) = 208 − 40 = 168 x = 17

b)

208 − 8. ( x + 4 ) = 2 3 .5 8. ( x + 4 ) = 208 − 40 = 168 x = 17

c) 

2018. ( 2019 − x ) = 0 x = 2019