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Ta có:\(\frac{x-1}{2013}+\frac{x-2}{2012}=\frac{x-3}{2011}+\frac{x-4}{2010}\Rightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1=\frac{x-3}{2011}-1+\frac{x-4}{2010}-1\)
\(\Rightarrow\frac{x-1-2013}{2013}+\frac{x-2-2012}{2012}=\frac{x-3-2011}{2011}+\frac{x-4-2010}{2010}\)
\(\Rightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}=\frac{x-2014}{2011}+\frac{x-2014}{2010}\)
\(\Rightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)
\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2013}< \frac{1}{2011};\frac{1}{2012}< \frac{1}{2010}\) nên \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}< 0\)
\(\Rightarrow x-2014=0\Rightarrow x=2014\)
a, ( 8x - 3 ) ( 3x + 2 ) - ( 4x + 7 ) ( x + 4 ) = ( 2x + 1 ) ( 5x - 1 )
( 24x2 + 16x - 9x - 6 ) - ( 4x2 - 16x - 7x + 28 ) = 10x2 - 2x + 5x -1
24x2 + 16x - 9x - 6 -4x2 - 16x - 7x - 10x2 + 2x - 5x = 6 + 28 - 1
10x2 -19x = 33
10x2 - 19x -33 = 0 \(\Leftrightarrow\)10x( x+ 3 ) + 11 ( x- 3 ) = 0
=> ( x- 3 ) ( 10x + 11 ) = 0\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-11}{10}\end{cases}}\)
b, 4( x - 1 ) ( x + 5 ) - ( x + 2 ) ( x + 5 ) = 3( x - 1 ) ( x + 2 )
4( x2 - 5x - x + 5 ) - ( x2 + 5x + 2x + 10 ) = 3( x2 + 2x - x - 2 )
4x2 - 20x - 4x + 20 - x2 - 5x - 2x - 10 = 3x2 + 6x - 3x - 6
( 4x2 - x2 ) + ( -20x - 4x - 5x - 2x ) + 20 - 10 = 3x2 + ( 6x - 3x ) - 6
3x2 - 31x - 3x2 - 3x = -6-10
-34x = -16
x = \(\frac{8}{17}\)
=> 1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010
=> -1/x+3 = 1/2010
=> 1/x+3 = 1/-2010
=> x+3 = -2010
=> x = -2010-3 = -2013
k mk nha
1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010
=> -1/x+3 = 1/2010
=> 1/x+3 = 1/-2010
=> x+3 = -2010
=> x = -2010-3 = -2013
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{-1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow-\left(x-3\right)=2010\)
\(\Rightarrow x=-2013\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow x=2007\)
f(x)=0 =>5x=0
hay x=0
f(x)=1 =>5x=1
=>x=1/5
f(x)=-5
=>5x=-5
=>x=-1
f(x)=2010
=>5x=2010
hay x=402
\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\).
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(=-\frac{1}{x+3}=\frac{1}{2010}\)
\(x=2010-\left(-3\right)=2013\)
a thấy: 1/2 = 6/12 ; 2/3=6/9 ; 3/4=6/8 Như vậy nếu khối 3 có 12 phần thì khối 4 có 9 phần, khối 5 có 8 phần.
Tổng số phần bằng nhau 12+9+8 = 29 (phần)
Từ 570 đến 590 có các số chia hết cho 5 là: 570;575;580;585;590 trong đó có 580 chia hết cho 29.
Số học sinh lớp 3: 580:29x12 = 240 (hs)
Số học sinh lớp 4: 580:29x9= 180 (hs)
Số học sinh lớp 5: 580:29x8 = 160 (hs)
Học sinh nào z bn? Đề mk có như z đâu?