Tính
\(\left(x+4\right)\left(x-4\right)-x^3+2018\)
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=\(\frac{1}{2}x\frac{2}{3}x...x\frac{2017}{2018}\)
=\(\frac{1}{2018}\)
bạn trừ ra là đc
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot....\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
\(=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)....\left(\frac{2017}{2018}\right)\)
\(=\frac{1.2.3....2017}{2.3.4...2018}\)
\(=\frac{1}{2018}\)
\(=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2017}{2018}\)
\(=\frac{1}{2018}\)
câu 1
A=-1
câu 2
\(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\left(x+1\right).\left(x+1\right)=16\)
\(\left(x+1\right)^2=16\)
\(\Rightarrow x+1=4\)
vậy x=3