\(=2^{1+2+3+4+...+100}\)

\(=2^{\frac{\left(100+1\right).100}{2}}\)

\(=2^{5050}\)

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{10^2}\right)\)

\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{99}{100}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)

\(=\frac{1.2.3...9}{2.3.4...10}.\frac{3.4.5...11}{2.3.4...10}\)

\(=\frac{1}{10}.\frac{11}{2}\)

\(=\frac{11}{20}\)

\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{10^2}\right)\)

\(=\frac{3}{4}.\frac{8}{9}....\frac{99}{100}\)

\(\Rightarrow\frac{3.8...99}{4.9....100}=\frac{22}{45}\)

a) [ 3x-1] + 4x - 3 = 7

 3x - 1  + 4x = 7 + 3 = 10

( 3 + 4 )x - 1 =10

7x - 1 = 10

7x=11

x=11/7

Câu tiếp theo làm tương tự nhé =))

) [ 3x-1] + 4x - 3 = 7

 3x - 1  + 4x = 7 + 3 = 10

( 3 + 4 )x - 1 =10

7x - 1 = 10

7x=11

x=11/7

Ta có:

\(1990^{10}+1990^9=1990^9.\left(1990+1\right)=1990^9.1991\)

\(1991^{10}=1991^9.1991\)

Vì 1990⁹<1991⁹ => 1990⁹.1991<1991⁹.1991

=>1990¹⁰+1990⁹<1991¹⁰

\(\left(a,b\right)=\frac{ab}{\left[a,b\right]}=\frac{180}{60}=3\)

Đặt \(a=3m,b=3n,\left(m,n\right)=1\).

Ta có: \(ab=3m.3n=9mn=180\Leftrightarrow mn=20\)

mà \(\left(m,n\right)=1\)nên ta có bảng giá trị sau: 

(2x+1)(6-x)= 12x-2x^2-x+6