Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
57 x 36 + 114 x 32 - 1999 - 2001
= 57 x 36 + 57 x 2 x 32 - ( 1999 + 2001 )
= 57 x 36 + 57 x 64 - 4000
= 57 x ( 36 + 64 ) - 4000
= 57 x 100 - 4000
= 5700 - 4000
= 1700
\(57\cdot36+114\cdot32-1999-2001\)
\(=57\cdot36+57\cdot64-4000\)
\(=57\cdot\left(36+64\right)-4000\)
\(=5700-4000=1700\)
Ta có:
1-1995/1997=1997/1997-1995/1997=1/1997
1-1999/2001=2001/2001-1999/2001=1/2001
Vì 1/1995 > 1/2001 nên 1999/2001<1995/1997
hãy t cho mk nha mn mk biết ơn mn nhìu lắm
\(\frac{2000\cdot2001-1001}{1999\cdot2002-999}=\frac{1999\cdot2001+2001-1001}{1999\cdot2001+1999-999}\)
\(=\frac{1999\cdot2001+1000}{1999\cdot2001+1000}=1\)
\(x\cdot\left(1990+2000+2001\right)=99\cdot7-99\cdot4-99-99\cdot2\)
\(\Leftrightarrow x\cdot\left(1999+2000+2001\right)=99\cdot\left(7-4-1-2\right)\)
\(\Leftrightarrow x\cdot\left(1999+2000+2001\right)=99\cdot0\)
\(\Leftrightarrow x\cdot\left(1999+2000+2001\right)=0\)
\(\Rightarrow x=0\)
\(x\times\left(1999+2000+2001\right)=99\times7-99\times4-99-99\times2\)
\(x\times\left(1999+2000+2001\right)=99\times\left(7-4-1-2\right)\)
\(x\times\left(1999+2000+2001\right)=99\times0\)
\(x\times\left(1999+2000+2001\right)=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
Ta có: 1 - 1995/1997 = 2/1997
1 - 1999/2001 = 2/2001
mà 2/1997 > 2/2001 nên 1995/1997 < 1999/2001
Ta có: 1 - 1995 / 1997 = 2/1997
1- 1999/2001= 2/2001
VÌ 2/1997 > 2/2001 Nên 1995/1997 < 1999/2001
Ta có:
\(1-\frac{1995}{1997}=\frac{2}{1997}>1-\frac{1999}{2001}=\frac{2}{2001}\)
Vậy nên:
\(\frac{1995}{1997}<\frac{1999}{2001}\)
Ta có:
1-1995/1997=2/1997>1-1999/2001
=>1995/1997<1999/2001
k cho mình nha
57x36+114x32-1999-2001
=57x36+57x2x32-(1999+2001)
=57x 36+57x64-4000
=57x(36+64)-4000
=57x100-4000
=5700-4000
=5300
57 × 36 + 114 × 32 - 1999 - 2001
= 57 × 36 + 57 × 2 × 32 - (1999 + 2001)
= 57 × 36 + 57 × 64 - 4000
= 57 × (36 + 64) - 4000
= 57 × 100 - 4000
= 5700 - 4000
= 1700