Chung minh rang : \(\dfrac{2}{x^2-1}+\dfrac{4}{x^2-4}+\dfrac{6}{x^2-9}+............+\dfrac{20}{x^2-100}\)chia het cho 11
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`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
Bài này có cần phải tính nhanh ko vậy bn?
Nếu ko thì lấy máy tính mà tính cũng đc mà
a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)
=>x=0
b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)
=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)
=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)
Đến đây thì bạn giải giống câu a
a) \(...\Rightarrow x.\left(2+5\right)=14\Rightarrow x.7=14\Rightarrow x=14:7=2\)
b) \(...\Rightarrow x.\left(9+1\right)=20\Rightarrow x.10=20\Rightarrow x=20:10=2\)
c) \(...\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=1999\Rightarrow x.\dfrac{3}{3}=1999\Rightarrow x=1999\)
d) \(...\Rightarrow11.x+22=5.x+40\Rightarrow11.x-5.x=40-22\Rightarrow6.x=18\Rightarrow x=18:6=3\)
e) \(...\Rightarrow11.x-66=4.x+11\Rightarrow11.x-4.x=11+66\Rightarrow7.x=77\Rightarrow x=77:7=11\)
f) \(...\Rightarrow\left(3.x-12\right):x=12-10\)
\(\Rightarrow3.x-12=2.x\)
\(\Rightarrow3.x-2.x=12\)
\(\Rightarrow x=12\)
g) \(...\Rightarrow\left(5.x+7\right):x=26-20\)
\(\Rightarrow5.x+7=6.x\)
\(\Rightarrow6.x-5.x=7\)
\(\Rightarrow x=7\)
h) \(...\Rightarrow x.\left(1999-1\right)=1999.\left(1997+1\right)\)
\(\Rightarrow x.1998=1999.1998\)
\(\Rightarrow x=1999.1998:1998\)
\(\Rightarrow x=1999\)
a, \(x\times\) 2 + \(x\times\) 5 = 14
\(x\) \(\times\) ( 2 + 5) = 14
\(x\) \(\times\) 7 = 14
\(x\) = 14: 7
\(x\) = 2
b, \(x\times9\) + \(x\)= 20
\(x\) \(\times\)( 9 + 1) = 20
\(x\) \(\times\) 10 = 20
\(x\) = 2
c, \(x\) : \(\dfrac{3}{2}\) + \(x\times\dfrac{1}{3}\) = 1999
\(x\times\) \(\dfrac{2}{3}\) + \(x\) \(\times\dfrac{1}{3}\) = 1999
\(x\times\) ( \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) = 1999
\(x\) = 1999
d, 11\(\times\)(\(x+2\)) = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) + 22 = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 40 - 22
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 18
11 \(\times\) \(x\) - 5 \(\times\) \(x\) = 18
\(x\) \(\times\) ( 11 - 5) = 18
\(x\) \(\times\) 6 = 18
\(x\) = 3