x;y là số j

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x-1}{2018}=\frac{3-y}{2019}=\frac{x-1+3-y}{2018+2019}=\frac{x-y+2}{4037}=\frac{4035+2}{4037}=\frac{4037}{4037}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2018}=1\\\frac{3-y}{2019}=1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-1=2018\\3-y=2019\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=2019\\y=-2016\end{cases}}\)

Vậy,.......

cảm ơn bn nha!

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

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

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

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

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

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

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

Lời giải:

Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$

$\Rightarrow x=2018a; y=2019a; z=2020a$

$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$

Mặt khác:

$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$

Từ $(1); (2)$ ta có đpcm.

\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

\(\dfrac{x}{2017}=\dfrac{y}{2018}=\dfrac{z}{2019}=k\\ \Rightarrow\left\{{}\begin{matrix}x=2017k\\y=2018k\\z=2019k\end{matrix}\right.\)

\(4\left(x-y\right)\left(y-z\right)=4\left(2017k-2018k\right)\left(2018k-2019k\right)=4\left(-k\right)\left(-k\right)=4k^2=\left(2k\right)^2=\left(2019k-2017k\right)^2=\left(z-x\right)^2\left(ĐPCM\right)\)

Áp dụng BĐT: \(\dfrac{a^n+b^n}{2}\ge\left(\dfrac{a+b}{2}\right)^n\Rightarrow a^n+b^2\ge2\left(\dfrac{a+b}{2}\right)^n\):

\(\left(1+\dfrac{x}{y}\right)^{2018}+\left(1+\dfrac{y}{x}\right)^{2018}\ge2\left(\dfrac{2+\dfrac{x}{y}+\dfrac{y}{x}}{2}\right)^{2018}\ge2\left(\dfrac{2+2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}}{2}\right)^{2018}=2^{2019}\)

Dấu "=" xảy ra khi \(x=y\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019