TL:

\(789302687-100=789302587\)

\(5639174929-649102=5638525827\)

_HT_

trên đầu á

số 0(chắc thế) xin T.I.C.H

Khó quá, mới lớp 2 thôi mà

\(\lim\limits_{x→0}\dfrac{\sqrt{x+1}-\sqrt[3]{x+1}}{x}\\ =\lim\limits_{x→0}\dfrac{\sqrt{x+1}-1}{x}+\lim\limits_{x→0}\dfrac{1-\sqrt[3]{x+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x+1-1}{x\left(\sqrt{x+1}+1\right)}+\lim\limits_{x\rightarrow0}\dfrac{1-\left(x+1\right)}{x\left(1+\sqrt[3]{x+1}+\left(\sqrt[3]{x+1}\right)^2\right)}\\ =\lim\limits_{x\rightarrow0}\dfrac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\dfrac{-1}{1+\sqrt[3]{x+1}+\left(\sqrt[3]{x+1}\right)^2}\)

\(=\dfrac{1}{\sqrt{1}+1}+\dfrac{-1}{1+\sqrt[3]{1}+\left(\sqrt[3]{1}\right)^2}\\ =\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)

lớp 6 mà hỏi toán lớp 11

k đi

hok tốt

@@@@@@@@@@@@@@@@@@

đó ai trl đc

`Answer:`

\(\text{lim }\frac{2.4^n+1}{1-4^{n+2}}\)

Chia cả tử và mẫu cho `4^n` được:

\(\text{lim }\frac{2.4^n+1}{1-4^{n+2}}\)

\(=\text{lim }\frac{\frac{2.4^n+1}{4^n}}{\frac{1-4^{n+2}}{4^n}}\)

\(=\text{lim }\frac{2+\frac{1}{4^n}}{\frac{1}{4^n}-4^2}\)

\(=\frac{2+0}{0-4^2}\)

\(=-\frac{1}{8}\)

báo cáo

B-á-o c-á-o-!

Nguyễn Thu Thủy Ngân

What ???