ta thấy vế thứ hai có kết quả bằng 0
=>(1999x1998+1998x1997)x0
chằng cần tìm kết quả mà =>B=0

bạn học nhà cô Hà dạy Toán cấp 2 đt đúng ko 

B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)

B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)

B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x 0

B = 0

#)Trả lời :

\(a,\frac{2}{3}:\frac{5}{7}.\frac{5}{7}:\frac{2}{3}+1934\)

\(=\left(\frac{2}{3}:\frac{2}{3}\right).\left(\frac{5}{7}:\frac{5}{7}\right)+1934\)

\(=1.1+1934\)

\(=1935\)

              #~Will~be~Pens~#

Trả lời : 

\(a,\text{ }\frac{2}{3}\text{ : }\frac{5}{7}\text{ x }\frac{5}{7}\text{ : }\frac{2}{3}+1934\)

\(=\left(\frac{2}{3}\text{ : }\frac{2}{3}\right)\text{ x }\left(\frac{5}{7}\text{ : }\frac{5}{7}\right)+1934\)

\(=1\text{ x }1+1934\)

\(=1935\)

B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)

B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)

B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x 0

B = 0

B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)

B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)

B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x 0

B = 0

b) \(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+...+\frac{87}{1000}+\frac{99}{1000}\)

\(=\frac{1+13+25+...+85+97}{1000}=\frac{\left(97+1\right).\left[\left(97-1\right):12+1\right]:2}{1000}\)

\(=\frac{49.9}{1000}=\frac{441}{1000}.\) ( Đề bài sai nhé bạn tử số : 1; 13; 25; 37; 49 ; 61; 73; 85 ; 97. )

Câu c:B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)

B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)

B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x 0

B = 0

B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)

B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)

B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)

B = (1999 x 1998 + 1998 x 1997) x 0

B = 0

=1x7/9

=7/9

9/7 Nếu 7/9 ở dưới mẫu

\(\frac{1999\cdot2001-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot\left(2000+1\right)-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot2000+1999-1}{1998+1999.2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot2000+1998}{1998+1999.2000}\cdot\frac{7}{5}=1\cdot\frac{7}{5}=\frac{7}{5}\)

thank you very much

=(3994002+1998+1997)x(\(\frac{2}{3}\)-\(1\frac{1}{3}\))

=3997997x\(\frac{-2}{3}\)

=-2665331,333