1/So sánh :
a/ \(\left(\frac{3}{8}\right)^5\&\left(\frac{5}{243}\right)^3\)
b/ \(M=\frac{10^{1992}+1}{10^{1991}+1};N=\frac{10^{1993}+1}{10^{1992}+1}\)
2/ Chứng minh rằng : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{50^2}<1\)
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\(A=x+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)
\(=5x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\)
\(=5x+2\)
\(B=5x\)
\(\Rightarrow A>B\)Với \(\forall\)\(x\)
#)Giải :
\(A=\left[x\right]+\left[1+\frac{1}{5}\right]+\left[x+\frac{2}{5}\right]+\left[x+\frac{3}{5}\right]+\left[x+\frac{4}{5}\right]\)
Thay x = 3,7 vào biểu thức, ta có :
\(A=\left[3,7\right]+\left[3,7+\frac{1}{5}\right]+\left[3,7+\frac{2}{5}\right]+\left[3,7+\frac{3}{5}\right]+\left[3,7+\frac{4}{5}\right]\)
\(A=\left[3,7+3,7+3,7+3,7+3,7\right]+\left[1+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right]\)
\(A=18,5+3\)
\(A=21,5\)
\(B=\left[5x\right]=\left[5\times3,7\right]=18,5\)
Vì 21,5 > 18,5 \(\Rightarrow A>B\)
Với n =1 thì A < 3. Vậy ta phải đi chứng minh A < 3
Giả sử A < 3 đúng với n = k. Ta có:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{k^2+3k}\right)< 3\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(\frac{k^2+3k+2}{k\left(k+3\right)}\right)\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}\)
Ta phải đi chứng minh A < 3 đúng với n = k +1 tức là phải chứng minh:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\left(1+\frac{2}{\left(k+1\right)^2+3\left(k+1\right)}\right)\) \(< 3+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\)
Ta sẽ có:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\left(1+\frac{2}{k^2+2k+1+3k+3}\right)\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\frac{k^2+5k+6}{k^2+5k+4}\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\) \(< 3+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\)
Vậy A đúng với n = k + 1 thì A đúng với n = k
Vậy A < 3 là điều phải chứng minh.
(Phương pháp quy nạp toán học)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)
\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)
Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)
b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)
\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)
Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)
Bài làm
Ta có: \(\left(-\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{4}\right)^2\right]^5=\left(\frac{1}{4}\right)^{10}\)
Mà \(2< 10\)
=> \(\left(\frac{1}{4}\right)^2< \left(\frac{1}{4}\right)^{10}\)
Hay \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)
Vậy \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)
# Học tốt #
Mình làm bài 2 nhé:
Ta có: \(\frac{1}{2^2}<\frac{1}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3^2}<\frac{1}{3\times4}=\frac{1}{3}-\frac{1}{4}\)
....
\(\frac{1}{50^2}<\frac{1}{50\times51}=\frac{1}{50}-\frac{1}{51}\)
Tổng các vế ta sẽ có \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{2}-\frac{1}{51}=\frac{49}{102}<1\)