tìm x biết
1600:(41-(2.x-5)3)=40
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\(720:\left[41-\left(2x-5\right)\right]=40\) \(3x+8x-30=25\)
\(41-\left(2x-5\right)=18\) \(3x+8x=55\)
\(2x-5=23\) \(11x=55\)
\(2x=28\) \(x=5\)
\(x=14\)
x= \(\left(40-1\right)\cdot2+1=79\)
Tổng bằng 1600 đúng rồi
\(1+3+5+...+x=1600\)
\(\Rightarrow\left[\frac{\left(x-1\right)}{2}+1\right]^2=1600\)
\(\Rightarrow\left[\frac{\left(x-1\right)}{2}+1\right]=40^2\)
\(\Rightarrow\frac{\left(x-1\right)}{2}+1=40\)
\(\Rightarrow\frac{\left(x-1\right)}{2}=40-1=39\)
\(\Rightarrow x-1=39.2=78\)
\(\Rightarrow x=78+1=79\)
Mà x là số lẻ \(\Rightarrow x=79\)
Vậy x = 79
1+3+5+...+x=1600
{(n-1)/(2)+1}^2=1600
(n-1)/(2)+1 = 40^2
(n-1)/(2)+1 = 40
(n-1)/(2) = 40-1
(n-1)/(2) = 39
n-1 = 39*2
n-1 = 78
n = 78 +1
n = 79
Do x lẻ nên x = 2.k + 1 (k thuộc N)
\(1+3+5+...+\left(2k+1\right)=1600\)
=> \(\left(2k+1+1\right).\left(\frac{2k+1-1}{2}+1\right):2=1600\)
=> \(\left(2k+2\right).\left(\frac{2k}{2}+1\right):2=1600\)
=> \(2.\left(k+1\right).\left(k+1\right):2=1600\)
=> \(\left(k+1\right)^2=1600=40^2\)
=> k + 1 = 40
=> k = 39
=> x = 2 x 39 + 1 = 79
a ) \(\left(4x+5\right)\div3-121\div11=4\)
\(\left(4x+5\right)\div3-11=4\)
\(\left(4x+5\right)\div3=4+11\)
\(\left(4x+5\right)\div3=15\)
\(\left(4x+5\right)=15\cdot3\)
\(4x+5=45\)
\(4x=45-5\)
\(4x=40\)
\(x=10\)
(4x + 5) : 3 - 121 : 11 = 4
=> (4x + 5) : 3 - 11 = 4
=> (4x + 5) : 3 = 15
=> 4x + 5 = 45
=> 4x = 40
=> x = 10
b) 1 + 3 + 5 + ... + x = 1600
=>[(x - 1) : 2 + 1] . (x + 1) : 2 = 1600
=> \(\left(\frac{x}{2}-\frac{1}{2}+1\right).\frac{x+1}{2}=1600\)
=> \(\frac{x+1}{2}.\frac{x+1}{2}=1600\)
=> \(\left(\frac{x+1}{2}\right)^2=1600\)
=> \(\frac{x+1}{2}=40\)
=> x + 1 = 80
=> x = 79
Tìm x biết :
1+3+5+...+x=1600 (x là số lẻ )
1+3+5+...+x=1600
{(n-1)/(2)+1}^2=1600
(n-1)/(2)+1 = 40^2
(n-1)/(2)+1 = 40
(n-1)/(2) = 40-1
(n-1)/(2) = 39
n-1 = 39*2
n-1 = 78
n = 78 +1
n = 79
1600 : ( 41 - ( 2x - 5 )3 ) = 40
41 - ( 2x - 5 ) = 1600 : 40
41 - ( 2x - 5 ) = 40
( 2x - 5)3 = 41 - 40
(2x - 5 )3 = 1
( 2x - 5 )3 = 13
2x - 5 = 1
2x = 1 + 5
2x = 6
x = 6 : 2
x = 3
Vậy x = 3