a) \(\left(\dfrac{1}{2}\right)^n\le10^{-9}\)\(\Leftrightarrow2^{-n}\le10^{-9}\)\(\Leftrightarrow-n\le log^{10^{-9}}_2\)\(\Leftrightarrow-n\le-9log^{10}_2\)\(\Leftrightarrow n\ge9log^{10}_2\)\(\Leftrightarrow n\ge30\).
Vậy \(n=30\).

b) \(3-\left(\dfrac{7}{5}\right)^n\le0\)

\(\Leftrightarrow-\left(\dfrac{7}{5}\right)^n\le-3\)

\(\Leftrightarrow\left(\dfrac{7}{5}\right)^n\ge3\)\(\Leftrightarrow n\ge log^3_{\dfrac{7}{5}}\)

\(\Rightarrow\)\(n\in\left\{4;5;6;7;...\right\}\Rightarrow n=4\)

c) \(1-\left(\dfrac{4}{5}\right)^n\ge0,97\)

\(\Leftrightarrow-\left(\dfrac{4}{5}\right)^n\ge-0,3\)

\(\Leftrightarrow\left(\dfrac{4}{5}\right)^n\le0,3\)\(\Leftrightarrow n\ge log^{0,3}_{\dfrac{4}{5}}\)

\(\Rightarrow n\in\left\{6;7;8;9...\right\}\Rightarrow n=6\)

d)\(\left(1+\dfrac{5}{100}\right)^n\ge2\)

\(\Leftrightarrow1,05^n\ge2\)

\(\Rightarrow n\in\left\{15;16;17;18;...\right\}\Rightarrow n=15\)

em mới lp 6 k biết trình bày kiểu lp 12

Không mất tính tổng quát, giả sử \(a\ge b\ge c\).

Khi đó: \(\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow ab+bc\ge ac+b^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{c}+1\ge\dfrac{a}{b}+\dfrac{b}{c}\\\dfrac{c}{a}+1\ge\dfrac{c}{b}+\dfrac{b}{a}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\le2+2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)

Vì \(1\le c\le a\le2\Rightarrow\left(\dfrac{a}{c}-2\right)\left(\dfrac{2a}{c}-1\right)\le0\)

\(\Leftrightarrow\dfrac{a}{c}+\dfrac{c}{a}\le\dfrac{5}{2}\)

\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\le7\)

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le10\)

Đẳng thức xảy ra khi \(a=b=2;c=1\) và các hoán vị.

Nguyễn Thanh Hằng,nguyen van tuan,Nguyễn Huy Tú,Ace Legona,... giúp mk vs

Lời giải:

a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n-1}-1\right)\left(\frac{1}{n}-1\right)\)

\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{-(n-2)}{n-1}.\frac{-(n-1)}{n}\)

\(=\frac{(-1)(-2)(-3)...[-(n-2)][-(n-1)]}{2.3.4...(n-1)n}\)

\(=\frac{(-1)^{n-1}(1.2.3....(n-2)(n-1))}{2.3.4...(n-1)n}=(-1)^{n-1}.\frac{1}{n}\)

b) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)

\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)

\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)

\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)

\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)

\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)

\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)

\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)

a/ \(\lim\limits\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}=\lim\limits\dfrac{\dfrac{\left(\dfrac{1}{3}\right)^{n+1}-1}{\dfrac{1}{3}-1}}{\dfrac{\left(\dfrac{1}{2}\right)^{n+1}-1}{\dfrac{1}{2}-1}}=\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}}=3\)

b/ \(\lim\limits\left(n^3+n\sqrt{n}-5\right)=+\infty-5=+\infty\)

1.

Trước hết bạn nhớ công thức:

$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)

Áp vào bài:

\(\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}\)

\(=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})\)

\(=1.\frac{1}{3}=\frac{1}{3}\)

2.

\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)

\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)

\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)

\(=x+a\) 

3.

Trước tiên ta có công thức:

$1^3+2^3+....+n^3=(1+2+3+...+n)^2=\frac{n^2(n+1)^2}{4}$
Chứng minh: https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/

Khi đó:

\(\lim \frac{1^3+2^3+...+n^3}{n^4}=\lim \frac{n^2(n+1)^2}{4n^4}\\ =\lim \frac{(n+1)^2}{4n^2}=\frac{1}{4}\lim (1+\frac{1}{n})^2=\frac{1}{4}.1=\frac{1}{4}\)

A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n⁴ + 4 = (n²+2)² - 4n² = (n²-2n+2)(n²+2n+2) = [(n-1)²+1][(x+1)²+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

a) \(\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^{n-1}}\)

\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n:\left(-\dfrac{5}{7}\right)}\)

\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n.\left(-\dfrac{7}{5}\right)}\)

\(=\dfrac{1}{\left(-\dfrac{7}{5}\right)}\)

\(=1.\left(-\dfrac{5}{7}\right)\)

\(=-\dfrac{5}{7}\)

b) \(\dfrac{\left(-\dfrac{1}{2}\right)^{2n}}{\left(-\dfrac{1}{2}\right)^n}\)

\(=\dfrac{\left(-\dfrac{1}{2}\right)^n.\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^n}\)

\(=\left(-\dfrac{1}{2}\right)^n\)