A=(1-1/2).(1-1/3).(1-1/4).(1-1/5). (....) . (1-1/2019)
các bạn ơi giúp mk vs ngày mai mk thi rồi
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: 3.4 = 2.6
\(\Rightarrow\frac{3}{2}=\frac{6}{4};\)\(\frac{3}{6}=\frac{2}{4};\)\(\frac{4}{2}=\frac{6}{3};\)\(\frac{4}{6}=\frac{2}{3}\)
b) Ta có: 14.15 = 10.21
\(\Rightarrow\frac{14}{10}=\frac{21}{15};\)\(\frac{14}{21}=\frac{10}{15};\)\(\frac{15}{10}=\frac{21}{14};\)\(\frac{15}{21}=\frac{10}{14}\)
c) Ta có: AB.CD = 2.3
\(\Rightarrow\frac{AB}{2}=\frac{3}{CD};\)\(\frac{AB}{3}=\frac{2}{CD};\)\(\frac{CD}{2}=\frac{3}{AB};\)\(\frac{CD}{3}=\frac{2}{AB}\)
d) Ta có: A.AB = 5.MN
\(\Rightarrow\frac{A}{5}=\frac{MN}{AB};\)\(\frac{A}{MN}=\frac{5}{AB};\)\(\frac{AB}{5}=\frac{MN}{A};\)\(\frac{AB}{MN}=\frac{5}{A}\)
a)3.4=2.6
=\(\frac{3}{2}=\frac{6}{4};\frac{3}{6}=\frac{2}{4};\frac{2}{3}=\frac{4}{6};\frac{6}{3}=\frac{4}{2}\)
b)14.15=10.21
=\(\frac{14}{10}=\frac{21}{15};\frac{14}{21}=\frac{10}{15};\frac{15}{10}=\frac{21}{14};\frac{15}{21}=\frac{10}{14}\)
c)AB.CD=2.3
=\(\frac{AB}{2}=\frac{3}{CD};\frac{AB}{3}=\frac{2}{CD};\frac{CD}{2}=\frac{3}{AB};\frac{CD}{3}=\frac{2}{AB}\)
D)A.AB=5.MN
=\(\frac{A}{5}=\frac{MN}{AB};\frac{A}{MN}=\frac{5}{AB};\frac{AB}{5}=\frac{MN}{A};\frac{AB}{MN}=\frac{5}{A}\)
xy - 2x - 3y = 1
=> (xy - 2x) - 3y + 6 = 1 + 6
=> x(y - 2) - 3(y - 2) = 7
=> (x-3)(y-2) = 7
=> x - 3; y - 2 thuộc Ư(7) = {-1; 1; -7; 7}
ta có bảng :
x-3 | -1 | 1 | -7 | 7 |
y-2 | -7 | 7 | -1 | 1 |
x | 2 | 4 | -4 | 10 |
y | -5 | 9 | 1 | 3 |
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\)
\(\frac{x}{4}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\frac{x+y+z}{4+5+6}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\) mà x + y + z = 45
\(\Rightarrow\frac{45}{15}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow3=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\hept{\begin{cases}x=3\cdot4=12\\y=3\cdot6=18\\z=3\cdot5=15\end{cases}}\)
\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)
\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)
\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))
Vậy B < 2
Ta có:
\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)
\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)
\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)
...
\(1+\frac{1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
=>
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)
\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)
Vậy B < 2
| 4x - 1 | + 4x =5
TH1: \(4x-1\ge0\)
\(4x-1+4x=5\)
\(4x+4x=5+1\)
\(8x=6\)
\(x=\frac{3}{4}\)( thỏa mãn \(4x-1\ge0\))
TH2: 4x - 1 <0
\(-\left(4x-1\right)+4x=5\)
\(-4x+1+4x=5\)
\(1=5\) ( vô lí)
Vậy \(x=\frac{3}{4}\)
1. A = 100
2. B = 2098
mik ko biết có đúng ko đâu nhé vì mình nhowfbanj làm cho rồi viết vô đây mà ahihi
\(\Leftrightarrow\left(1-3x\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow1-3x=-5\)
\(\Leftrightarrow3x=1-\left(-5\right)\)
\(\Leftrightarrow3x=6\)
\(\Rightarrow x=2\)
a, ta có (-5)3=-125
=>1-3x=-5
3x=1+5
3x=6
x=2
b,34-x=27
81-x=27
x=54
nha
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2019}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2018}{2019}\)
\(A=\frac{1.2.3.....2018}{2.3.4....2019}\)
\(A=\frac{1}{2019}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2018}{2019}\)
\(A=\frac{1}{2019}\)
Bn ơi tk cho mk nha!!!!!!!!!!!