Than khảo! Tự viết lại đề nhé!!
=> \(3.4^x.4^2-5.4^x:4=748\)
=> \(4^x.(48-\frac{5}{4})\) = 748
=>\(4^x. \frac{187}{4}=748\)
=> \(4^x=16=4^2\)
=> x = 2
Than khảo! Tự viết lại đề nhé!!
=> \(3.4^x.4^2-5.4^x:4=748\)
=> \(4^x.(48-\frac{5}{4})\) = 748
=>\(4^x. \frac{187}{4}=748\)
=> \(4^x=16=4^2\)
=> x = 2
(-1/2)^3+(-1/2)^2+(-1/2)^1+(-1/2)^0
C=\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}< -\frac{1}{2}\)
D=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)< -\frac{1}{2}\)
tính: a)(-1)x(-1)^2x(-1)^3x(-1)^4x...x(-1)^9x(-1)^10
b)[1/100-1^2]x[1/100-(1/2)^2]x[1/100-(1/3)^2]x...x[1/100-(1/20)^2]
1.So sánh: A=\(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\) và 1:
1.So sánh: A=\(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\) và 1:
So sánh
\(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}và\frac{1}{2^2.3.5^2.7}\)
rút gọn biểu thức :
N = 1 + \(\left(\dfrac{1}{2}\right)\) + \(\left(\dfrac{1}{2}\right)^2\) + \(\left(\dfrac{1}{2}\right)^3\) + ... + \(\left(\dfrac{1}{2}\right)^{100}\)
Tính
\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{n^2}\right)\left(n\in N,n\ge2\right)\)
So sánh A với \(-\dfrac{1}{2}\) biết:
A=\(\left(\dfrac{1}{2^2}-1\right)\)\(\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(1-\dfrac{1}{100^2}\right)\)