45.32+210-23.30
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số lớn là :
(95,75+45,32):2=70,535
Đ/S : 70,535
. la phẩy hả
\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}-\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+...+\dfrac{7}{23.30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{23}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{77}{240}-\dfrac{7}{15}\right)=\dfrac{1}{7}.\left(-\dfrac{7}{48}\right)=-\dfrac{1}{48}\)
1/3.10+1/10.17+......+1/73.80 - 1/2.9 - 1/9.16 - 1/16.23 - 1/23.30
= (7/3.10+7/10.17+......+7/73.80) : 7 - (7/2.9 + 7/9.16 + 7/16.23 + 7/23.30) : 7
= (1/3-1/10+1/10-1/17+...+1/73-1/80) : 7 - (1/2-1/9+1/9-1/16+1/16-1/23+1/23-1/30) : 7
=(1/3-1/80) : 7 - (1/2-1/30) : 7
= 77/240 : 7 - 7/15 : 7
=11/240 - 1/15
= -1/48
Nhấn đúng cho mk nha!!!!!!!!!!!!!
a.45.32+45.66+45.2
= 45.(32+66+2)
=45.100=4500
b.(2013+2014+2015).(20+10).12-18.20
=(2013+2014+2015).(30.12-(12+6).20)
=(2013+2014+2015).(30*12-12.20-6.20)
=(2013+2014+2015).(12.(30-20)-6.20)
=(2013+2014+2015).(12.10-6.20)
=(2013+2014+2015).(120-120)
=(2013+2014+2015).0
=0
c.(12+12+12.88).45+12.12
=(12.(1+1+88)).45+12.10
=12.90.45+120
=12.9.10.5.9+120
=1080.5.9+120
=5400.9+120
=48600+120
=48720
nho k cho to nhe
đề sai thì phải
\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)
\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)
\(A=\frac{101}{120}\)
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)
\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)
Biến đổi 3980= 2*1990, nhóm nhân tử được A=1
Nhóm nhân tử được B=1
Vậy A=B
\(A=\frac{1990.1991+3980}{1993.1990}\) \(B=\frac{23.55+23.45}{23.70+23.30}\)
\(A=\frac{1990.1991+1990.2}{1993.1990}\) \(B=\frac{23.\left(55+45\right)}{23.\left(70+30\right)}\)
\(A=\frac{1990.\left(1991+2\right)}{1993.1990}\) \(B=\frac{23.100}{23.100}=1\)
\(A=\frac{1990.1993}{1993.1990}=1\)
\(\Rightarrow A=B\)
Nếu phân số thứ 2 là \(\frac{1}{10.17}\) thì làm như vậy nè
\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}\right)-\left(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+\frac{1}{23.30}\right)\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}\right)-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
= \(\frac{1}{7}.\frac{77}{240}-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{30}\right)=\frac{1}{7}.\frac{77}{240}-\frac{1}{7}.\frac{7}{15}\)
= \(\frac{11}{240}-\frac{1}{15}\)
= \(-\frac{1}{48}\)
=10000
học tốt
=1024.9+1024-8.30
=1024.(9-8).30
=1024.30
=30720