a: \(8A=8+8^2+...+8^8\)
\(\Leftrightarrow7A=8^8-1\)
hay \(A=\dfrac{8^8-1}{7}\)
b: \(B=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)}{3^2-1}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)}{8}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)}{8}=\dfrac{3^{16}-1}{8}\)
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
So sánh \(A=3^{32}-1\) và \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Tìm \(x\)
a) \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
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=\(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)-2^{32}\)
PHƯƠNG PHÁP THÊM BỚT HẠNH TỬ VÀ PHƯƠNG PHÁP ĐỔI BIẾN :
a) \(4x^4+1\)
b) \(x^8+4\)
c) \(x^4+x^2+1\)
d) \(x^7+x^5+1\)
e) \(x^7+x^5-1\)
f) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
g) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
h) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)
i) \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)
GIải phương trình sau:
a) \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
b) \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
Tính giá trị biểu thức:
a) \(\left(x-10\right)^2-x.\left(x+8\right)với\)\(x=0,98\)
b) \(x^3-9x^2+27.x-27\) với x =5
c) \(6x.\left(2x-7\right)-\left(3x-5\right).\left(4x+7\right)\) tại x = \(-2\)
Bài 1 : Chứng minh rằng với mọi số nguyên n
a) \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\) chia hết cho 5
b)\(n\left(n+5\right)-\left(n-3\right)\left(n+2\right)\)chia hết cho 6
c)\(\left(n-1\right)\left(n+1\right)-\left(n-7\right)\left(n-5\right)\)chia hết cho 12
Bài 2:
Tìm x biết : \(\left(4x+3_{^{ }}\right)^3+\left(5-7x\right)^3+\left(3x-8\right)^3=0\)
Cho\(A=2^{32}+1;B=\left(2+1\right)\left(2^2+1\right)\left(2^8+1\right)\left(2^{16+1}\right)\)
so sánh A Và B
