Câu 1:
a: ĐKXĐ: x>=1
\(\sqrt[3]{2-x}=1-\sqrt{x-1}\)
=>\(\sqrt[3]{2-x}-1+\sqrt{x-1}=0\)
=>\(\frac{2-x-1}{\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{2-x}+1}+\frac{x-1}{\sqrt{x-1}}=0\)
=>\(\frac{1-x}{\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{2-x}+1}+\frac{x-1}{\sqrt{x-1}}=0\)
=>\(\left(x-1\right)\left(\frac{-1}{\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{2-x}+1}+\frac{1}{\sqrt{x-1}}\right)=0\)
=>x-1=0
=>x=1(nhận)
b: Ta có công thức: \(1-\frac{2}{n\left(n+1\right)}\)
\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(S=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot\ldots\cdot\left(1-\frac{2}{2020\cdot2021}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(2020+2\right)\left(2020-1\right)}{2020\left(2020+1\right)}\)
\(=\frac{4\cdot5\cdot\ldots\cdot2022}{3\cdot4\cdot\ldots\cdot2021}\cdot\frac{1\cdot2\cdot\ldots\cdot2019}{2\cdot3\cdot\ldots\cdot2020}=\frac{2022}{3}\cdot\frac{1}{2020}=\frac{1011}{1010\cdot3}=\frac{337}{1010}\)
