Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x+2x+3x+4x+...+2011x = 2012.2013
\(\Rightarrow\) x(1+2+3+4+...+2011) = 4050156
\(\Rightarrow\) x.2023066 = 4050156
\(\Rightarrow\) x = 4026/2011
\(x+2x+3x+....+2001x=2012.2013\)
\(\Leftrightarrow x\left(1+2+3+...+2001\right)=2012.2013\)
\(\Leftrightarrow x.\frac{\left(2011+1\right).2011}{2}=2012.2013\)
\(\Leftrightarrow x=\frac{2012.2013.2}{\left(2011+1\right).2011}\)
\(\Leftrightarrow x=\frac{2.2013}{2011}\)
Cứ 1 số hạng lại kèm theo 1x
Số số hạng từ 1 đến 2011 là:
( 2011 - 1 ) : 1 + 1 = 2011 ( số hạng )
Do đó có 2011x
Ta có:\(x+2x+3x+4x+...+2011x=2012.2013\)
\(2011x=2012.2013\)
\(x=\frac{2012.2013}{2011}\)
a) (1+2+3+....+2011)x=2012.2013
<=>\(\frac{2011.2012}{2}\)x=2012.2013
<=>x=4026/2011
b)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
<=>\(\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
<=>x=2012
c)dùng công thức \(\frac{2}{\left(2x-1\right)\left(2x+1\right)}=\frac{1}{2x-1}-\frac{1}{2x+1}\)
ta được 1-1/2x+1=2 giải ra được x
ok
\(x+2x+3x+...+2011x=2012.1013\)
\(\dfrac{2011\left(2011+1\right)}{2}x=2012.2013\)
\(x=2012.2013.\dfrac{2}{2011.2012}\)
\(x=\dfrac{4026}{2011}\)
a) [2011(2011+1):2].x = 2012.2013 => x = (2012.2013):(2011.1006) b) =>(2x-1)^m -x^m = 0 => (2x-1)^m = x^m =>2x-1 = x => x = 1
46:
\(A=\dfrac{2x^2\left(3x^2-2x+1\right)}{2x^2}-\left(3x^2-x-6x+2\right)\)
\(=3x^2-2x+1-3x^2+7x-2=5x-1\)
Khi x=-0,2 thì A=-1-1=-2
45:
a: \(=\dfrac{-5x^6}{3x^2}=-\dfrac{5}{3}x^4\)
c: \(=\dfrac{2x\left(2x^2-\dfrac{3}{2}x+1\right)}{2x}=2x^2-\dfrac{3}{2}x+1\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
moi hok lop 6 thoi
X+2X+3X+4X+...+2011X=2012.2013
<=> X(1+2+3+4+......+2011)=2012.2013
<=>2023066X=4050156
<=>X=4050156:2023066=\(\frac{4026}{2011}\)
Vậy x=\(\frac{4026}{2011}\)