Lời giải:

Đặt mẫu số của $B$ là $M$.

Từ \(2018x^3=2019y^3=2020z^3\)

\(\Rightarrow \sqrt[3]{2018}x=\sqrt[3]{2019}y=\sqrt[3]{2020}z=\frac{\sqrt[3]{2018}}{\frac{1}{x}}=\frac{\sqrt[3]{2019}}{\frac{1}{y}}=\frac{\sqrt[3]{2020}}{\frac{1}{z}}=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)

\(=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{8}=\frac{M}{8}\)

\(\Rightarrow \left\{\begin{matrix} x=\frac{M}{8\sqrt[3]{2018}}\\ y=\frac{M}{8\sqrt[3]{2019}}\\ z=\frac{M}{8\sqrt[3]{2020}}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2018x^2=\frac{\sqrt[3]{2018}M^2}{64}\\ 2019y^2=\frac{\sqrt[3]{2019}M^2}{64}\\ 2020z^2=\frac{\sqrt[3]{2020}M^2}{64}\end{matrix}\right.\)

\(\Rightarrow 2018x^2+2019y^2+2020z^2=\frac{M^2(\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020})}{64}=\frac{M^3}{64}\)

\(\Rightarrow B=\frac{\sqrt[3]{\frac{M^3}{64}}}{M}=\frac{M}{4M}=\frac{1}{4}\)

\(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)+z^3-3xyz-3xy\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-xz=0\end{cases}}\)

Mà \(x,y,z>0\Rightarrow x+y+z\ne0\)

\(\Rightarrow x^2+y^2+z^2-xy-xz-yz=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow}x=y=z\)

Thay vào biểu thức A ta được : 

\(A=\frac{2018x-2019x+2020x}{\sqrt[3]{x^3}}=\frac{2019x}{x}=2019\)

Vậy ...

đây nha bạn

nhanh nhanh các bạn cần gấp

\(\dfrac{y+z+t-2020x}{x}=\dfrac{z+t+x-2020y}{y}=\dfrac{t+x+y-2020z}{z}=\dfrac{x+y+z-2020t}{t}=\dfrac{-2017\left(x+y+z+t\right)}{x+y+z+t}=-2017\\ \Leftrightarrow\left\{{}\begin{matrix}y+z+t-2020x=-2017x\\z+t+x-2020y=-2017y\\t+x+y-2020z=-2017z\\x+y+z-2020t=-2017t\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y+z+t=2x\\x+y+z+t=2y\\x+y+z+t=2z\\x+y+z+t=2t\end{matrix}\right.\\ \Leftrightarrow x=y=z=t=\dfrac{x+y+z+t}{2}=1010\\ \Leftrightarrow A=1010\left(2019-2020+2021-2022\right)=1010\left(-2\right)=-2020\)

2x+y+3z=6(1)3x+4y−3z=4(2){2x+y+3z=6(1)3x+4y−3z=4(2)

Từ hệ phương điều kiện, ta có:

Lấy (1) + (2) ta được: 5x+5y= 10 ⇒⇒ x+y=2 ⇔⇔ y=2-x (3)

từ(1) ta suy ra y=6-3z-2x thế biểu thức vào phương trình (2) , ta được :

-5x-15z=-20 ⇔⇔ x+3z=4 ⇔⇔ z =43−x343−x3 (4)

thay (4) và (2) vào P ta được :

P= 2x+3y-4z = 2x +3.(2-x)- 4.(43−x343−x3) =2x+6-3x-163+4x3=x3+23163+4x3=x3+23

⇒⇒Min P ⇔⇔ x3x3 đạt GTNN mà 3>0 cố định ⇒⇒ Min P⇔⇔ x đạt GTNN

Mà x >= 0, x là số thực nên Min P = 2323 ,dấu "=" xảy ra khi và chỉ khi :

x=0

Ta có x + y = 2 ⇒⇒ y=2 ; z = 43−x343−x3 ⇒⇒ z =43

\(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\Rightarrow ab+bc+ca=2020\)

BĐT trở thành:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020.2021}{abc}\)

\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}+a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020.2021}{abc}\)

\(\Leftrightarrow a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020^2}{abc}\)

Ta có: \(\sqrt{2020+a^2}=\sqrt{ab+bc+ca+a^2}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{1}{2}\left(2a+b+c\right)\)

Tương tự:...

\(\Rightarrow\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le2\left(a+b+c\right)\)

\(\Rightarrow a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le3\left(a+b+c\right)\)

Nên ta chỉ cần chứng minh:

\(3\left(a+b+c\right)\le\dfrac{2020^2}{abc}=\dfrac{\left(ab+bc+ca\right)^2}{abc}\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\) (hiển nhiên đúng)

Dấu "=" xảy ra khi \(a=b=c\) hay \(x=y=z\)

Bổ xung \(\ge2\sqrt{2020}\)

Chỗ cuối là 2019z² nha

\(1010x^2+2xy+1010y^2+1009\left(x^2+y^2\right)\ge1010x^2+2xy+1010y^2+2018xy\)

\(=1010\left(x^2+2xy+y^2\right)=1010\left(x+y\right)^2\)

\(\Rightarrow\sqrt{2019x^2+2xy+2019y^2}\ge\sqrt{1010}\left(x+y\right)\)

Làm tương tự và cộng lại

\(\Rightarrow VT\ge1010\left(x+y+y+z+z+x\right)=\sqrt{1010}.2\sqrt{2}=2\sqrt{2020}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{\sqrt{2}}{3}\)