Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=1990\cdot1994\)
\(A=\left(1992-2\right)\cdot1994\)
\(A=1992\cdot1994-2\cdot1994\)
\(B=1992\cdot1992\)
\(B=\left(1994-2\right)\cdot1992\)
\(B=1994\cdot1992-2.1992\)
Ta có : B > A ( 2 x 1992 < 2 x 1994 )
\(\frac{1991.1992.1993.1994.995}{1990.1991.1992.1993.997}=\frac{1994.995}{1990.997}=\frac{2.1}{2.1}=\frac{2}{2}=1\)
đáng ra là toán lớp 6 đó nhưng mik thích đặt toán lớp 5 :)
A = \(\dfrac{10^{1990}+1}{10^{1991}+1}\) ⇒ 10A = \(\dfrac{10^{1991}+10}{10^{1991}+1}\) = \(1+\dfrac{9}{10^{1991}+1}\)
B = \(\dfrac{10^{1991}+10}{10^{1992}+1}\) ⇒ 10B = \(\dfrac{10^{1992}+10}{10^{1992}+1}\) = 1 + \(\dfrac{9}{10^{1992}+1}\)
Vì \(\dfrac{9}{10^{1991}+1}\) > \(\dfrac{9}{10^{1992}+1}\)
10A > 10B => A > B
Ta có:
\(A=1993\times1993\)
\(A=1993^2\)
Áp dụng HĐT \(a^2-b^2=\left(a-b\right)\left(a+b\right)\), ta có:
\(B=1992\times1994\)
\(B=\left(1993-1\right)\left(1993+1\right)\)
\(B=1993^2-1^2\)
\(B=1993^2-1\)
Mà 19932 > 19932 - 1
\(\Rightarrow A>B\)
A = \(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) +\(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\) + \(\dfrac{4}{9\times11\times13}\)
A = \(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+...+\(\dfrac{1}{9\times11}\)-\(\dfrac{1}{11\times13}\)
A = \(\dfrac{1}{1\times3}\) - \(\dfrac{1}{11\times13}\)
A = \(\dfrac{1}{3}-\dfrac{1}{143}\)
A = \(\dfrac{140}{429}\)
Bài 2:
A = \(\dfrac{1991}{1990}\) x \(\dfrac{1992}{1991}\) x \(\dfrac{1993}{1992}\) x \(\dfrac{1994}{1993}\) x \(\dfrac{1995}{997}\)
A = \(\dfrac{1994\times1995}{1990\times997}\)
A = \(\dfrac{997\times2\times5\times399}{5\times2\times199\times997}\)
A = \(\dfrac{399}{199}\)
a)
\(=\dfrac{4.4.25+4.11.4.25}{29.2.48+71.2.48}=\dfrac{400+400.11}{96.29+71.96}=\dfrac{400\left(1+11\right)}{96\left(29+71\right)}=\dfrac{400.12}{96.100}=\dfrac{2.2.100.12}{12.2.2.2.100}=\dfrac{1}{2}\)
1990*1994 =3968060
1992 * 1992 = 3968064
vậy 1990*1994 < 1992 * 1992