-có ai có công thức chung để tính tổng các số lũy thừa 2 như thế này ko ạ
-tại sao 6(12+22+...+n2)=6.\(\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Viết dưới dạng lũy thừa của 1 số nguyên
a)\(12^3:\left(3^{-4}.64\right)\) b) \(\left(\dfrac{3}{7}\right)^5.\left(\dfrac{7}{3}\right)^{-1}.\left(\dfrac{5}{3}\right)^6:\left(\dfrac{343}{625}\right)^{-2}\)c) \(5^4.125.\left(2,5\right)^{-5}.0,04\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
Viết dưới dạng lũy thừa của 1 số nguyên
a)\(12^3:\left(3^{-4}.64\right)\) b) \(\left(\dfrac{3}{7}\right)^5.\left(\dfrac{7}{3}\right)^{-1}.\left(\dfrac{5}{3}\right)^6:\left(\dfrac{343}{625}\right)^{-2}\)c) \(5^4.125.\left(2,5\right)^{-5}.0,04\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
Viết dưới dạng lũy thừa của 1 số nguyên
a)\(12^3:\left(3^{-4}.64\right)\) b) \(\left(\dfrac{3}{7}\right)^5.\left(\dfrac{7}{3}\right)^{-1}.\left(\dfrac{5}{3}\right)^6:\left(\dfrac{343}{625}\right)^{-2}\)c) \(5^4.125.\left(2,5\right)^{-5}.0,04\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
Viết dưới dạng lũy thừa của 1 số nguyên
a)\(12^3:\left(3^{-4}.64\right)\) b) \(\left(\dfrac{3}{7}\right)^5.\left(\dfrac{7}{3}\right)^{-1}.\left(\dfrac{5}{3}\right)^6:\left(\dfrac{343}{625}\right)^{-2}\)c) \(5^4.125.\left(2,5\right)^{-5}.0,04\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
Tìm giới hạn các dãy số sau
a) \(lim\dfrac{2^n+6^n-4^{n-1}}{3^n+6^{n+1}}\)
b) \(lim\dfrac{1+3+5+...+\left(2n+1\right)}{3n^2+4}\)
c) \(lim\dfrac{1+2+3+...+n}{n^2-3}\)
d) \(lim\left[\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right]\)
e) \(lim\left[\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right]\)
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
Viết dưới dạng lũy thừa của một số nguyên:
a/\(12^3:\left(3^{-4}.64\right)\)
b/\(\left(\dfrac{3}{7}\right)^5.\left(\dfrac{7}{3}\right)^{-1}.\left(\dfrac{5}{3}\right)^6:\left(\dfrac{343}{625}\right)^{-2}\)
c/\(5^4.125.\left(2,5\right)^{-5}.0,04\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6\cdot\dfrac{5}{3}\cdot\dfrac{3}{7}:\left(\dfrac{7^3}{5^4}\right)^{-2}\)
\(=\left(\dfrac{5}{7}\right)^6\cdot\dfrac{5}{7}\cdot\left(\dfrac{5}{7}\right)^6\cdot5^2\)
\(=\left(\dfrac{5}{7}\right)^{13}\cdot5^2\)
c: \(=5^4\cdot2.5^{-5}\cdot125\cdot0.04\)
\(=5^4\cdot5\cdot\left(\dfrac{5}{2}\right)^{-5}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
Bài 1: Viết các biểu thức sau đưa dạng lũy thừa của một số hữu tỉ
a)4.64.28
b)128.27
c)4.27:\(\left(3^{11}.\dfrac{1}{9}\right)\)
Bài 2: Tính
a)\(\left(\dfrac{1}{2}\right)^3\).4+\(\dfrac{3}{4}\)
b)46.\(\left(\dfrac{1}{2}\right)\)12
c)\(\left(\dfrac{1}{2}\right)^5\)- 1,52
d)\(\dfrac{14^{16}.35^7}{10^9.7^{22}}\)
e)\(\dfrac{4^{20}-2^{20}}{6^{20}-5^{20}}\)
Giúp mình với nhé! Mình tick cho ! Cảm ơn mọi người !
tính các giới hạn sau:
a) lim (3n2+n2-1)
b)lim \(\dfrac{n^3+3n+1}{2n-n^3}\)
c) lim \(\dfrac{-2n^3+3n+1}{n-n^2}\)
d) lim \(\left(n+\sqrt{n^2-2n}\right)\)
e) lim \(\left(2n-3.2^n+1\right)\)
f) lim \(\left(\sqrt{4n^2-n}-2n\right)\)
g) lim \(\left(\sqrt{n^2+3n-1}-\sqrt[3]{n^3-n}\right)\)
a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
a) lim \(\left(-3n^3+n^2-1\right)\)
minh le oi ban dao mau so cua ban len cho tu uong roi thay vi tri cua mau thanh n3 +2n
quan sát kĩ ta có mấy công thức về lũy thừa đúng trong 6 công thức sau :
với x bất kì còn m , n ∈ Z :
\(x^m.x^n=x^{m+n}\)(1) \(x^m:x^n=x^{m-n}\)(2) (x\(^m\))\(^n\) = \(x^{m.n}\)(3)
\(\left(x.y\right)^n=x^ny^n\)(4) \(\left(\dfrac{x}{y}\right)^n=\dfrac{x^n}{y^n}\) (5) (với y ≠ 0)
\(x^n=x.x...x\) (6) (n thừa số x,x ≠ 0)