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![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
#)Giải :
Bài 1 :
\(A=\frac{15}{6.16}+\frac{15}{16.26}+\frac{15}{26.36}\)
\(A=3\left(\frac{5}{6.16}+\frac{5}{16.26}+\frac{5}{26.36}\right)\)
\(A=3\left(\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{26}+\frac{1}{26}-\frac{1}{36}\right)\)
\(A=3\left(\frac{1}{6}-\frac{1}{36}\right)\)
\(A=3.\frac{5}{36}\)
\(A=\frac{15}{36}\)
\(B=\frac{33}{6.16}-\frac{63}{16.26}+\frac{93}{26.36}\)
\(B=\frac{3.11}{6.16}-\frac{3.21}{16.26}+\frac{3.31}{26.36}\)
\(B=\frac{1}{2.16}.11+\frac{1}{2.16}.\frac{3.21}{13}+\frac{3.31}{26.36}\)
\(B=\frac{1}{32}.\frac{143-63}{13}+\frac{3.31}{26.12.3}\)
\(B=\frac{1}{32}.\frac{80}{13}+\frac{31}{36.12}\)
\(B=\frac{5}{26}+\frac{31}{26.12}=\frac{60+31}{26.12}=\frac{91}{26.12}=\frac{13.7}{13.2.12}=\frac{7}{24}\)
Bài 2 :
a)\(28.157-28.267+28.10\)
\(=28\left(157-267+10\right)\)
\(=28.-100\)
\(=-2800\)
b)\(\left(-5\right).4.\left(-2\right).23.\left(-25\right)\)
\(=\left[4.\left(-25\right)\right].\left[\left(-5\right).2\right].23\)
\(=\left(-100\right).\left(-10\right).23\)
\(=23000\)
2)
28 . 157 - 28 .267 + 28.10
=28.( 157 - 267 + 10 )
= 28 . -100
= -2800
3)
[(-2).(-5)]. 4 .-25.23
= 10 . -100 .23
=-23000
còn câu 1 mik suy nghĩ đã
![](https://rs.olm.vn/images/avt/0.png?1311)
=( 21999+21999.25): (21990+9)
=21999.(1+25): 21999
=21999.(1+25): 21999
=21999. (1+25): 21999
=1.(1+32)
=1.33
=33
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi tổng của phép tính trên là A
\(A=15+2^4+2^5+2^6+...+2^{2022}\)
\(\Rightarrow2A=2+2^2+...+2^{2023}\)
\(\Rightarrow A=2^{2023}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=(3-5)+(7-9)+.....(319-321)
=(-2)+(-2)+....+(-2)
=(-2) x 160
=(-320)
Ta nhóm 2 số vào một nhóm ta được 160 nhóm có giá trị là -2 ta được -2.160=-320
Nhỏ bấm đúng ch mình nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (1/4)3 x (1/8)2
= [(1/2)2]3 x [(1/2)3]2
= (1/2)6 x (1/2)6
= (1/2)12
b) 42 x 32: 23
= (22)2 x 25: 23
= 24 x 25: 23
= 24 x 22
= 26
c) 25 x 53 x 1/625 x 53
= 52x 53 x (1/5)4 x 53
= (1/5)4 x 58
= 1/54 x 58 (giải thích nếu ko hiểu: (1/5)4= 14/54= 1/54)
= 58/54
= 54
d) 56 x 1/20 x 22 x 32 : 125
= 56/20 x (2x3)2 : 53
= 56/ (5x4) x 62: 53
= 55/4 x 62/53 (62/53 là dạng phân số, bản chất vẫn là lấy 62 chia 53)
= 55 x 62/ 4x 53 (nhân phân số: tử nhân tử, mẫu nhân mẫu)
= 52x 62/ 22 (chia 55 cho 53 ra 52)
= 302/ 22
= 152
*Kiến thức áp dụng:
amx an = am+n
am: an= am-n
(am)n = am x n
am x bm = (a x b)m
![](https://rs.olm.vn/images/avt/0.png?1311)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)
\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)
\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)
\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)
\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)
mik xinloi ạ câu 50 mik viết sai đề nên mong bạn giải lại giúp ạ <3
\(a,=\left(-4\right)\\ b,=\left(-70\right)\\ c,2002+\left[45-5^2\right]=2002+\left[45-25\right]\\ =2002+20=2022\)
-4
-133
6026