Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 = (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
#)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~#
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
(1999 x 1998 + 1998 x 1997) x (1 + 1/2 : 3/2 - 4/3)
= (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x 0
= 0
(1999 x 1998 + 1998 x 1997) x (1 + 1/2 : 3/2 - 4/3)
= (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x 0
= 0
\(\frac{1999\cdot2001-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)
\(=\frac{1999\cdot\left(2000+1\right)-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)
\(=\frac{1999\cdot2000+1999-1}{1998+1999.2000}\cdot\frac{7}{5}\)
\(=\frac{1999\cdot2000+1998}{1998+1999.2000}\cdot\frac{7}{5}=1\cdot\frac{7}{5}=\frac{7}{5}\)
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
bn ở tỉnh nào vậy
B=0 tự giải nhé anh lớn lắm rồi nha em
\(B=\left(1999\cdot1998+1998\cdot1997\right)\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
\(B=\left(1999\cdot1998+1998\cdot1997\right)\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
\(B=\left(1999\cdot1998+1998\cdot1997\right)\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(B=\left(1999\cdot1998+1998\cdot1997\right)\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(B=\left(1999\cdot1998+1998\cdot1997\right)\cdot0\)
\(B=0\)
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