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Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
a)
4^5+4^5+4^5+4^5/3^5+3^5+3^5 =4^5.4/3^5.3
=4^6/3^6
tương tự ta có phân số còn lại là
6^5.6/=2^5.2
=6^6/2^6
ta có 4^6/3^6+6^6/2^6 ta rút gọn 4^6 với 2^6 và 6^6 với 3^6 được 2/1+2/1=2^2
suy ra số đó là 2
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(-\frac{5}{6}\right)-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{7}{8}+\frac{6}{7}\)
\(=\frac{7}{8}+\frac{6}{7}=\frac{49}{56}+\frac{48}{56}=\frac{49+48}{56}=\frac{97}{56}\)
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
\(\left(\frac{5}{4}-\frac{5}{3}-\frac{5}{2}\right)\left(x+5\right)=\frac{5}{6}+\frac{5}{4}-\frac{5}{8}\)
=> \(5\left(\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)\left(x+5\right)=\frac{-5}{2}\left(-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}\right)\)
=> \(5\left(x+5\right)=-\frac{5}{2}\)
=> 5x + 25 = - 2,5
=> 5x = -27,5
=> x = -5,5
(5/4-5/3-5/2).(x+5)=(5/6+5/4-5/8)
(5/4-5/3-5/2).(x+5)=35/24
-35/12.(x+5)=35/24
(x+5)=35/24:-35/12
(x+5)=35/24.-12/35
(x+5)=-1/2
x=-1/2-5
x=-5,5
vậy x=-5,5
a) 12 ( 2x - 6 ) - 43= 53
12 ( 2x - 6 ) - 43= 125
12 ( 2x - 6 ) =125+43
12 ( 2x - 6 ) =168
2x - 6=168:12
2x - 6=14
2x=14+6
2x=20
x=20:2
x=10
b) 7 ( 25 - 3x ) = 34 - 25
7 ( 25 - 3x ) = 49
25 - 3x =49:7
25 - 3x =7
3x=25-7
3x=18
x=18:3
x=6
c) ( 3x - 5 ) + 34 + 60 = 53
( 3x - 5 ) + 81 + 1 = 125
( 3x - 5 ) + 81=125-1
( 3x - 5 ) + 81=124
3x - 5=124-81
3x - 5=43
3x=43+5
3x=48
x=48:3
x=16
d) 3x + 2x ( 23 . 5 - 32 . 4 ) + 52 = 44
3x + 2x ( 8 . 5 - 9 . 4 ) + 25 = 256
3x + 2x.4 + 25 = 256
3x + 2x.4=256-25
3x + 2x.4=231
(3+2).x.4=231
5x.4=231
5x=231:4
5x=57,75
x=57,75:5
x=11,55
e) 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
Bài 1:
a) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=4\frac{5}{7}-1\frac{3}{4}\)
\(=\frac{33}{7}-\frac{7}{4}\)
\(=\frac{132}{28}-\frac{49}{28}=\frac{83}{28}\)
b) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=4\frac{5}{9}-2\frac{3}{4}\)
\(=\frac{41}{9}-\frac{11}{4}\)
\(=\frac{164}{36}-\frac{99}{36}=\frac{65}{36}\)
c) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\cdot\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=\frac{-3}{5}\cdot2=-\frac{6}{5}\)
d) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
Theo đề bài ta có:
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\frac{4^5\cdot4}{3^5\cdot3}\cdot\frac{6^5\cdot6}{2^5\cdot2}=2x\)
\(\Leftrightarrow\frac{4^6}{3^6}\cdot\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\left(\frac{4}{3}\cdot\frac{6}{2}\right)^6=2x\)
\(\Leftrightarrow4^6=2x\)
\(\Leftrightarrow2^{12}=2x\)
\(\Rightarrow x=2^{11}=2048\)