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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{204}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{203}{204}\)
\(=\frac{1}{204}\)
\(\text{Sửa đề }\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times....\times\left(1-\frac{1}{203}\right)\times\left(1-\frac{1}{204}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times....\times\frac{202}{203}\times\frac{203}{204}\)
\(=\frac{1\times2\times3\times...\times202\times203}{2\times3\times4\times...\times203\times204}\)
\(=\frac{1}{204}\)
\(a,\dfrac{7}{12}-\left(x+\dfrac{7}{10}\right):\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{10}:\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{12}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{5}{4}+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{11}{6}\)
\(\Leftrightarrow x=\dfrac{7}{12}-\dfrac{11}{6}\)
\(\Leftrightarrow\dfrac{-5}{4}\)
a,(X-4).(x-7)<0 => X-4<0,X-7>0 hoac X-4>0,X-7<0 => -7<x<-4 hoac 7>x>4
<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x
2x^2-8x-17-2x^2-2=0
-8x-19=0
x=-19/8
Ta có :
31.57.x - 57.14.x=85.57
=>x.(31.57-57.14)=85.57
=>x.[ 57.(31-14)] = 85.57
=>x. 57. 17 = 85.57
=>x.17=85
=>x=5
----------------------HOK TỐT---------------------------
\(2^{x+1}=4=2^2\Rightarrow x+1=2\Rightarrow x=1\)
\(2^{x-1}=4=2^2\Rightarrow x-1=2\Rightarrow x=3\)
\(2^x.4=128=4.32=4.2^5\Rightarrow x=5\)
\(3^{x-2}.9=81=9.9=9.3^2\Rightarrow x-2=2\Rightarrow x=4\)
\(\frac{3^{x+1}}{9}=3\Rightarrow3^{x+1}=27=3^3\Rightarrow x=2\)
\(41-2^x=9\Rightarrow2^x=41-9=32=2^5\Rightarrow x=5\)
Quá 20h rồi không biết được ko
\(\Leftrightarrow x+x+...+x+1+2+3+...+2008=2008.2009\)
\(\Leftrightarrow x.2008+\frac{\left(1+2008\right).2008}{2}=2008.2009\)
\(\Leftrightarrow x.2008=2008.2009-\frac{2008.2009}{2}\)
\(\Leftrightarrow x.2008=\frac{2008.2009}{2}\)
\(x=\frac{2009}{2}\)
\(9x-5x+3x=56\)
<=> \(7x=56\)
<=> \(x=8\)
Vậy...
\(9x-x+3x=165\)
<=> \(11x=165\)
<=> \(x=15\)
Vậy....
\(7x+8x-x=168\)
<=> \(14x=168\)
<=> \(x=12\)
Vậy...
\(\frac{5}{8}x-\frac{1}{3}x-\frac{1}{6}x=15\)
<=> \(\frac{1}{8}x=15\)
<=> \(x=120\)
Vậy...
Dấu . là dấu nhân nha
---
a, 9 . x - 5 . x + 3 . x = 56
x . (9 - 5 + 3) = 56
x . 7 = 56
x = 56 : 7
x = 8
b, 9 . x - x + 3 . x = 165
x . (9 - 1 + 3) = 165
x . 11 = 165
x = 165 : 11
x = 15
c, 7 . x + 8 . x - x = 168
x . (7 + 8 - 1) = 168
x . 14 = 168
x = 168 : 14
x = 12
d, 5/8 . x - 1/3 . x - 1/6 . x = 15
x . (5/8 - 1/3 - 1/6) = 15
x . 1/8 = 15
x = 15 : 1/8
x = 120
`3 xx x xx (x-2)=0`
`@TH1 : 3 xx x = 0 => x = 0 :3 = x=0`
`@TH2 : x-2=0 => x = 0 + 2 = x=2`
Vậy `x= {0;2}`
_____________________________
`(2 xx x -2)xx x = 0`
`@TH1:`
`2 xx x - 2=0`
`=> 2x =2`
`=> x = 2 : 2
`x=1`
`@TH2:
`x=0`
Vậy `x = {1;0}`
__________________________________
`(2xx x -4)xx(x-10)=0`
`@TH1:`
`2 xx x - 4 = 0`
`2 xx x = 0 + 4`
`2 xx x = 4`
`=> x = 2`
`@TH2:`
`x-10=0`
`x=0+10`
`x=10`
Vậy`x = {2;10}`
`3.x.(x-2)=0`
`@TH1:3x=0=>x=0`
`@TH2:x-2=0=>x=2`
___________________________________
`(2x-2)x=0`
`@TH1:2x-2=0=>2x=2=>x=1`
`@TH2:x=0`
________________________________________
`(2x-4)(x-10)=0`
@TH1:2x-4=0=>2x=4=>x=2`
`@TH2:x-10=0=>x=10`