\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
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=> x+x+1+x+2+...+x+30 = 1240
=> 31x + (1+2+...+30)=1240
=> 31x + 465 = 1240
=> 31x = 1240 - 465 = 775
=> x =775 : 31 = 25
Vậy x = 25
a, (19x+2.52) : 14 = (13-8)2 - 42
(19x + 2.25) : 14 = 52 - 42
(19x + 50) : 14 = 25 - 16
(19x + 50) : 14 = 9
19 x + 50 = 9.14
19x + 50 = 126
19x = 126 - 50
19x = 76
x = 76 : 19
x = 4
vậy____
b) x + (x + 1) + (x + 2) + (x + 3)+.....+(x+30) = 1240
(x+x+x+...+x) + (1+2+3+...+30) = 1240
31x + 465 = 1240
31x = 1240 - 465
31x = 775
x = 775 : 31
x = 25
vậy____
c) |x + 7| = 20 + 5.(-3)
|x + 7| = 20 + (-15)
|x + 7| = 5
\(\Rightarrow\orbr{\begin{cases}x+7=5\\x+7=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5-7\\x=-5-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=-12\end{cases}}\)
vậy_____
a) https://olm.vn/hoi-dap/question/1286785.html
b)
SSH là :
( x - 1 ) : 1 + 1 = x
Tổng là :
( x + 1 ) . x : 2 = 210
x ( x + 1 ) = 420
mà x và x + 1 là 2 số liên tiếp => 420 = 20 x 21 => x = 20
Vậy,............
\(a,\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=5^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=9\)
\(\Leftrightarrow19x+50=126\)
\(\Leftrightarrow19x=76\Leftrightarrow x=4\)
b) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1240
x + x + 1 + x + 2 + ... + x + 30 = 1240
( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240
Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )
Tổng là : ( 30 + 1 ) . 30 : 2 = 465
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
Vậy........
a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
x+(x+1)+(x+2)+...+(x+30)=1240
=> (x+x+...+x) + (1+2+3...+30) = 1240
=> 31x + 465 = 1240
=> 31x = 1240 - 465 = 775
=> x = 775 : 31
=> x = 25
= ( x+x+x+x+x+...+x ) + ( 1+2+3+..+30 ) =1240
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25