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(16/7 + -11/3) + (6/5 + -2/7) + 37/15
mn giải giúp mik với ạ. Với có hc lazi ko ạ, ko thì mik sorry.
![](https://rs.olm.vn/images/avt/0.png?1311)
mik có hc lazi nè
link của của mik https://lazi.vn/user/bach.bach17
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
=> \(5A=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\)
Lấy 5A trừ A theo vế ta có :
5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\right)\)
4A = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt B = \(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
=> 5B = \(1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
Lấy 5B trừ B ta có :
=> 5B - B = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)\)
=> 4B =\(1-\frac{1}{5^{11}}\)
=> B = \(\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó 4A = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{1}{5^{12}}\)
=> A = \(\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{1}{5^{12}.4}\right)< \frac{1}{16}\left(\text{ĐPCM}\right)\)
cậu ơi , mình quên không ghi 1 dữ liệu ạ
n thuộc N
V ậy có cần phải chỉnh sửa ở trong bài làm không ạ?????
![](https://rs.olm.vn/images/avt/0.png?1311)
Tách 1+2-3-4+5-6-8+...+97-98-99-100 với 101
A= 1 + 2 - 3 -4 + 5 + 6 -7 -8 + ... +97 +98 -99 -100 ( có: ( 100 - 1 ) : 1 + 1 = 100 )
A= ( 1 +2 - 3 - 4 ) + ( 5 + 6 - 7 -8 ) + ... ( 97 + 98 - 99 +100 ) ( có 100 : 4 = 25 cặp )
A= - 4 + -4 + -4 + ... + -4 ( có 25 số hạng )
A= ( -4 ) . 25
A= -100 + 101
A=1
học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\dfrac{-7}{4}\cdot\dfrac{2}{9}=\dfrac{-14}{36}=\dfrac{-7}{18}\)
2: \(=\dfrac{12}{13}\cdot\dfrac{26}{5}=\dfrac{24}{5}\)
3: \(=\dfrac{20}{11}\cdot\dfrac{55}{21}=\dfrac{100}{21}\)
4: \(=\dfrac{-40}{240}=\dfrac{-1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, x/7 =-4/14`
`=> 14x=7.(-4)`
`=>14x=-28`
`=>x=-28:14`
`=>x=-2`
`b,x/2=-2/-x`
`=>x/2=2/x`
`=>x.x=2.2`
`=>x^2=4`
`=>x= +-2`
`c,(x-1)/5=5/(x-1)`
`=>(x-1)^2 = 5.5`
`=>(x-1)^2=25`
`=>(x-1)^2=5^2`
\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
`d,x+3/2=-12/16`
`=>x=-12/16 -3/2`
`=>x= -12/16 - 24/16`
`=>x= -36/16`
`=>x=-9/4`
![](https://rs.olm.vn/images/avt/0.png?1311)
a)9.x + 1=73
9x=73-1
9x=72
x=72:9
x=8
b)2.x - 5 = -17 - 12
2x-5=-29
2x=-29+5
2x=-24
x=-24:2
x=-12
c)10 - x - 5 = - 5 - 7 -11
10-x-5=-12-11
10-x-5=-23
10-x=-23+5
10-x=18
x=10-18
x=-8
d)(-9) . x + 3 = (-2) . (-7) +16
-9x+3=14+16
-9x+3=30
-9x=30-3
-9x=27
x=27:(-9)
x=-3
(-12) . x - 34 =2
-12x=2+34
-12x=36
x=36:(-12)
x=-3
(-11).x + 9 =130
-11x=130-9
-11x=121
x=121:(-11)
x=11
(-5) .x + 5 = (-15) .(-4) -12
-5x+5=60-12
-5x+5=48
-5x=48-5
-5x=43
x=43:(-5)
x=-8,6
IxI -3=0
|x|=3
=>x=+3
(7 - IxI).(2.x - 4) =0
*7-|x|=0 * 2x-4=0
|x|=7 2x=4
=>x=+7 x=4:2
x=2
280-(x-140):35 =270
(x-140):35=280-270
(x-140):35=10
x-140=10.35
x-140=350
x=350+140
x=490
(1900 - 2.x ) : 35- 32 =16
1900-2x=(16+32).35
1900-2x=1680
2x=1900-1680
2x=220
x=220:2
x=110
720 :[41-(2x -5 )] =23 .5
720:[41-(2x-5)]=40
41-(2x-5)=720:40
41-(2x-5)=18
2x-5=41-18
2x-5=23
2x=23+5
2x=28
x=28:2
x=14
(x - 5).(x2 - 4 ) =0
* x-5=0 * x2-4=0
x=0+5 x2=4
x=5 x2=22
=> x=+2
Sửa đề P = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)CM P < 5/16
=> 5P = \(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\)
Lấy 5P trừ P theo vế ta có
5P - P = \(\left(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\right)-\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\right)\)
4P = \(1+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{5^2}-\frac{2}{5^2}\right)+...+\left(\frac{11}{5^{10}}-\frac{10}{5^{10}}\right)-\frac{11}{5^{11}}\)
4P = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
Đặt Q = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
=> 5Q \(=5+1+\frac{1}{5}+...+\frac{1}{5^9}\)
Lấy 5Q trừ Q theo vế ta có
5Q - Q = \(\left(5+1+\frac{1}{5}+...+\frac{1}{5^9}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\right)\)
4Q \(=5-\frac{1}{5^{10}}\)
=> Q\(=\frac{5}{4}-\frac{1}{5^{10}.4}\)
Khi đó 4P = \(\frac{5}{4}-\frac{1}{5^{10}.4}-\frac{11}{5^{11}}\)
=> P = \(\frac{5}{16}-\frac{1}{5^{10}.16}-\frac{11}{5^{11}.4}\)
\(=\frac{5}{16}-\frac{1}{5^{10}}\left(\frac{1}{16}-\frac{11}{5.4}\right)< \)\(\frac{5}{16}\)
Bài làm:
Ta có: \(\frac{1}{5^2}+\frac{2}{5^2}+\frac{3}{5^2}+...+\frac{11}{5^2}\)
\(=\frac{1+2+3+...+11}{5^2}=\frac{\left(1+11\right).11:2}{5^2}=\frac{66}{25}>1>\frac{1}{16}\)
\(\Rightarrow P>\frac{1}{16}\)
=> Đề sai