Phép nhân và phép chia các đa thức
Các câu hỏi tương tự
(3+1)*(3^2+1)*(3^4+1)*(3^8+1)*(3^16+1)*(3^32+1)
So sánh \(A=3^{32}-1\) và \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
(3+1)(3\(^2\) + 1)(3\(^4\) + 1)(3\(^8\) + 1)(3\(^{16}\) + 1)(3\(^{32}\) + 1)
Rút gọn biểu thức :
A= (38+1).(34+1).(32+1)(3+1)
B= 12.(52+1).(54+1).(58+1)...(532+1)
Rút gọn biểu thức
3(2^2 +1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
bài 3 tìm x
a)(2x+3)^2-(2x+1)(2x-1)=32
b)(4x+3)(4x-3)-(4x-5)^2=46
c)(x+4)^2+(x+3)(x-3)-5(x+1)(x-1)=16
d)92x-1)^2+(x+3^2-5(x+7)(x-7)=0
e)25x^2-9=0
10. Tính các tổng:
a) \(A=1+8+8^2+...+8^7\)
b) \(B=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
tính nhanh
a) A=\(2018^2-2017\cdot2019\)
b) B=\(9^8\cdot2^8-\left(18^4-1\right)\cdot\left(18^4+1\right)\)
c) C=\(163^2+74\cdot163+37^2\)
d) D=\(\dfrac{2018^3-1}{2018^2+2019}\)
e) E=\(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)-2^{32}\)
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)