K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
AL
help mik với
4
TC
25 tháng 3 2022
a,2/3 + 4/-5 x 20/16 =-1/3
b,(1/3+4/6).(2/7+9/14)=13/14
c,(2/3 - 3/4).(1/2- -3/5)=-11/120
TG
25 tháng 3 2022
a,2/3 + 4/-5 x 20/16
=2/3-1
=-1/3
b,(1/3+4/6).(2/7+9/14)
=1.13/14
=13/14
c,(2/3 - 3/4).(1/2- -3/5)
=-1/12.-1/10
=1/120
6 tháng 8 2018
b) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{20}}\)
\(\Rightarrow A=1-\frac{1}{2^{20}}< 1\left(đpcm\right)\)
c) ta có: \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{10}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{7}{10}\) ( có 7 số 1/10)
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{9}{19}\) ( có 9 số 1/19)
\(\Rightarrow\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{7}{10}+\frac{9}{10}=1\frac{33}{190}>1\)
=> đ p c m
6 tháng 8 2018
d) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
=> đ p c m
e) ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{7^2}< \frac{1}{6.7};\frac{1}{8^2}< \frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{7^2}+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}< 1\)
=> đ p c m
câu a mk ko bk, xl bn nhìu! :(
C
17 tháng 6 2017
\(a\left(\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\frac{2}{5}.y=\frac{1}{3}\)
\(y=\frac{1}{3}:\frac{2}{5}\)
\(y=\frac{5}{6}\)
\(b,\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}:\frac{10}{11}\)
\(y=\frac{22}{30}\)
28 tháng 8 2020
1/ (x+1)(y+2) =5
Do x;y thuộc N nên x+1 ; y+2 cũng thuộc N
\(TH1:\Leftrightarrow\hept{\begin{cases}x+1=1\\y+2=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=1-1\\y=5-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\y=3\end{cases}}}\\\)
\(TH2:\Leftrightarrow\hept{\begin{cases}x+1=5\\y+2=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=5-1\\y=1-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=-1\end{cases}}}\)
| x | 0 | 4 |
| y | 3 | -1 |
mà x;y\(\in\)N nên x;y=0;3
Các bài khác bạn làm tương tự nha! (vì mk viết rất chậm )
2 tháng 5 2019
Đặt \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}=A\)
ta có :\(\frac{1}{2^2}=\frac{1}{2\cdot2}=\frac{1}{4}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(...\)
\(\frac{1}{1990^2}=\frac{1}{1990\cdot1990}< \frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2\cdot3}+...+\frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\left(ĐPCM\right)\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
hk tốt #
2 tháng 5 2019
Ta có \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1990^2}< \frac{1}{1989.1990}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
\(\Rightarrow\)Bài toán được chứng minh
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
a) \(4\dfrac{3}{8}+5\dfrac{2}{3}\)
\(=\dfrac{35}{8}+\dfrac{17}{3}\)
\(=\dfrac{105}{24}+\dfrac{136}{24}\)
\(=\dfrac{241}{24}\)
b) \(2\dfrac{3}{8}+1\dfrac{1}{4}+3\dfrac{6}{7}\)
\(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}\)
\(=\dfrac{29}{8}+\dfrac{27}{7}\)
\(=\dfrac{419}{56}\)
c) \(2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}\)
\(=\dfrac{19}{8}-\dfrac{5}{4}+\dfrac{16}{3}\)
\(=\dfrac{9}{8}+\dfrac{16}{3}\)
\(=\dfrac{155}{24}\)
d) \(\left(\dfrac{5}{2}+\dfrac{1}{3}\right):\left(1-\dfrac{1}{2}\right)\)
\(=\dfrac{17}{6}:\dfrac{1}{2}\)
\(=\dfrac{17}{6}\cdot2\)
\(=\dfrac{17}{3}\)
e) \(\left(\dfrac{5}{2}-\dfrac{1}{3}\right)\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{13}{6}\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{39}{4}-\dfrac{6}{7}\)
\(=\dfrac{249}{28}\)
9 tháng 8 2023
a: =4+3/8+5+2/3
=9+9/24+16/24
=9+25/24
=216/24+25/24=241/24
b: \(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}=\dfrac{19+10}{8}+\dfrac{27}{7}\)
=27/7+29/8
=419/56
c: =2+3/8-1-1/4+5+1/3
=6+3/8-1/4+1/3
=6+3/8+1/12
=144/24+9/24+2/24
=155/24
d: =(15/6+2/6):1/2
=17/6*2
=17/3
e: =(15/6-2/6)*9/2-6/7
=13/6*9/2-6/7
=117/12-6/7
=249/28
ND
Nguyễn Đức Trí
VIP
8 tháng 8 2023
a) \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{37}{3}-\dfrac{17}{2}=\dfrac{74}{6}-\dfrac{51}{6}=\dfrac{23}{6}\)
b) \(3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+\dfrac{13}{6}.6=\dfrac{23}{6}+\dfrac{78}{6}=\dfrac{101}{6}\)
c) \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}=-\dfrac{92}{14}=-\dfrac{46}{7}\)
d) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}.\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)
8 tháng 8 2023
a. \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{23}{6}\)
\(b.3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+13=\dfrac{101}{6}\)
\(c.3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{20}{7}\)
d \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{11}\)
\(=\dfrac{101}{22}\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+\dfrac{1}{2^5}-\dfrac{1}{2^6}\)
\(2A=1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{2^5}\)
\(2A+A=\left(1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{2^5}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+\dfrac{1}{2^5}-\dfrac{1}{2^6}\right)\)
\(3A=1-\dfrac{1}{2^6}\Leftrightarrow A=\dfrac{1}{3}-\dfrac{1}{3.2^6}< \dfrac{1}{3}\left(đpcm\right)\)