\(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)....\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
So sánh B với \(\dfrac{11}{21}\)
Cho \(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{2020^2}-1\right).\) So sánh B và\(\dfrac{1}{2}\)
HELP ME!
So sánh : \(\left(\dfrac{1}{2}\right)^{12}\)và \(\left(\dfrac{1}{3}\right)^9\)
Tính: \(B=\left(1-\dfrac{1}{2^2}\right).\left(1-\dfrac{1}{3^2}\right).\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\) rồi so sánh với \(\dfrac{1}{2}\)
cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}-1\right).....\left(\dfrac{1}{10}-1\right)\). so sánh a với \(-\dfrac{1}{9}\)
So sánh 2 biểu thức sau A=\(\dfrac{y^2\left(x+1\right)+\left(x+1\right)}{y^2+1}\) và B=\(\dfrac{y^2\left(x-1\right)+2x-x}{y^2+2}\)
Bài 1 : So sánh :
\(\left(\frac{1}{16}\right)^{200}\)và \(\left(\frac{1}{2}\right)^{1000}\)
Bài 2 : Tính :
\(\left(6^9.2^{10}+12^{10}\right):\left(2^{19}.27^3+15.4^9.9^4\right)\)
P =1+\(\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
Cho \(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
So sánh \(A\) với \(\dfrac{-11}{21}\)
Giải chi tiết dùm mik nha. Thankss
