\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=\frac{3}{2}\Leftrightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=\frac{3}{2}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=\frac{1}{2}\)

\(\Leftrightarrow\frac{a+b+1}{ab}=\frac{1}{2}\Leftrightarrow2\left(a+b+1\right)=ab\Leftrightarrow2a+2b+2-ab=0\)

\(\Leftrightarrow2a-ab-4+2b+6=0\Leftrightarrow a\left(2-b\right)-2\left(2-b\right)=-6\)

\(\Leftrightarrow\left(a-2\right)\left(2-b\right)=-6\)

Đến đây chắc dễ rồi

@@@@@

Ta có:

\(\left(\frac{a}{b}\right)^3=\frac{1}{1000}=\left(\frac{1}{10}\right)^3\)

\(\Rightarrow\frac{a}{b}=\frac{1}{10}\Rightarrow\frac{a}{1}=\frac{b}{10}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{a}{1}=\frac{b}{10}=\frac{b-a}{10-1}=\frac{36}{9}=4\)

\(\Rightarrow\begin{cases}a=4.1=4\\b=4.10=40\end{cases}\)

Vậy a = 4; b = 10

Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{19}{20}\)

\(=\frac{1.2.3.....19}{2.3.4.....20}\)

\(=\frac{1}{20}>\frac{1}{21}\)

SORRY EM MỚI LỚP 6

a) 

\(A=1+2+3+...+1000\) và \(B=1.2.3....11\)

\(A=\frac{1000.1001}{2}=500500\) và \(B=11!=39916800\)

\(\Rightarrow A>B\)

b) 

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).......\left(1-\frac{1}{20}\right)\) và \(\frac{1}{21}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{19}{20}=\frac{1.2.3....19}{2.3.4....20}=\frac{1}{20}\) và \(\frac{1}{21}\)

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).......\left(1-\frac{1}{20}\right)\)\(>\frac{1}{21}\)

Có : (a/b)^3 = 1/1000 =(1/10)^3

<=> a/b = 1/10

<=> a = b/10 

Khi đó : b - b/10 = 36

<=> 9/10 . b = 36

<=> b = 36 : 9/10 = 40

<=> a = b/10 = 40/10 = 4

Vậy a= 4; b= 40

(\(\frac{a}{b}\))³=\(\frac{1}{1000}\)=(\(\frac{1}{10}\))³ => a/b=1/10 hay b=10a

=> 10a-9a=36 <=> 9a=36 => a=4; b=36+4=40

ĐS: a=4; b=40

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

b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)

c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)

\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)

\(A-B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}-\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)\)

\(A-B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)+\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)-\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)\)

\(A-B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)+\left[\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)-\left(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{2019}\right)\right]\)

\(A-B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)-0\)

\(A-B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\)

\(\text{Thay }A-B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\text{ ta có : }\)

\(\left(A-B-1\right)^{1000}=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}-1\right)^{1000}\)

\(=\left(1-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)^{1000}\)

\(=\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999}\right)^{1000}\)

Áp dụng BĐT Cosi:

\(\frac{a^4}{\left(a+2\right)\left(b+2\right)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}>=4\sqrt[4]{\frac{\left(a+2\right)\left(b+2\right)}{27.27.9}.\frac{a^4}{\left(a+2\right)\left(b+2\right)}}...\)

\(>=\frac{4}{9}a\)

Tương tự

\(=>VT>=\frac{4}{9}\left(a+b+c\right)-\frac{3}{9}-2\left(\frac{a+2}{9}+\frac{b+2}{9}+\frac{c+2}{9}\right)=\frac{1}{3}.\)

Dấu "="xảy ra khi a=b=c=1