b)x+ (x + 1) + (x + 3) + (x + 5) + . . . + (x + 29) = 2025
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a) (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 2025
(x + x + x + x + x) + (1 + 2 + 3 + 4 + 5) = 2025
5x + 15 = 2025
5x = 2025 - 15
5x = 2010
x = 2010 : 5
x = 402
b) 5 * x - x = 2020
5 * x - x * 1 = 2020
x * (5 - 1) = 2020
x * 4 = 2020
x = 2020 : 4
x = 505
mong bạn tick
a) ( x + 1 ) + ( x + 2) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 2025
\(\left(x+x+x+x+x\right)+\left(1+2+3+4+5\right)=2025\)
\(5x+15=2025\)
\(5x=2025-15\)
\(5x=2010\)
\(x=2010:5\)
\(x=402\).
2 x A = 1 - \(\dfrac{1}{2027}\)
\(A=\dfrac{1013}{2027}\)
(1+2+3+4+5+6+7+8+9+...............................+2016+2025) x (24,2 - 24,2) = (1 + 2 +3+4+5+6+7+8+9+...............................+2016+2025) x 0 = 0
b) \(2025^x=9^4\cdot5^4\)
\(\left(45^2\right)^x=\left(9\cdot5\right)^4\)
\(45^{2x}=45^4\)
\(\Rightarrow2x=4\)
\(x=4:2\)
\(x=2\)
Vậy x = 2
=))
Với giả thiết x2 - 4x + 1 = 0 thì\(B=x^5-3x^4-3x^3+6x^2-20x+2025=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2020=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2020=\left(x^3+x^2+5\right)\left(x^2-4x+1\right)+2020=2020\)
a) \(\frac{x+4}{x+3}< 1\)
\(\Leftrightarrow\frac{x+4}{x+3}-1< 0\)
\(\Leftrightarrow\frac{x+4-x-3}{x+3}< 0\)
\(\Leftrightarrow\frac{1}{x+3}< 0\)
\(\Leftrightarrow x+3< 0\)
\(\Leftrightarrow x< -3\)
Vậy \(x< -3\)
b) \(\frac{x+3}{x+4}>1\)
\(\Leftrightarrow\frac{x+3}{x+4}-1>0\)
\(\Leftrightarrow\frac{x+3-x-4}{x+4}>0\)
\(\Leftrightarrow-\frac{1}{x+4}>0\)
\(\Leftrightarrow x+4< 0\)
\(\Leftrightarrow x< -4\)
Vậy \(x< -4\)
c) \(\frac{x+3}{2010}+\frac{x+2}{2011}+\frac{x+1}{2012}+\frac{x+2025}{4}=0\)
\(\Leftrightarrow\left(\frac{x+3}{2010}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2025}{4}-3\right)=0\)
\(\Leftrightarrow\frac{x+2013}{2010}+\frac{x+2013}{2011}+\frac{x+2013}{2012}+\frac{x+2013}{4}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+2013=0\) (Vì \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\ne0\))
\(\Leftrightarrow x=-2013\)
Vậy \(x=-2013\)
Nhớ tick đó ✔✔✔
a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
a) X = {2012 ; 2016 ; 2020 ; 2024}
b)
y + 3 ⋮ 3 => y ⋮ 3
Mà: y ∈ {0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6} và y ≠ 0 nên y ∈ {3 ; 6}.
Vậy số cần tìm là 312 ; 612.
Vậy số cần tìm là 120 ; 126.
x+( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ............. + ( x + 29 ) = 2025
( x + x + x + ... + x ) + ( 1 + 3 + 5 + .. + 29 ) = 2025
x * 10 + 100 = 2025
x * 10 = 2025 - 100
x * 10 = 2025
x = 2025 : 10
x = 202,5