ĐKXĐ: x<>-2
\(B=\dfrac{3x-5}{x+2}=\dfrac{3x+6-11}{x+2}\)
\(=3-\dfrac{11}{x+2}\)
Để B là số nguyên âm lớn nhất thì \(\dfrac{11}{x+2}\) min
=>x+2=1
=>x=-1
ừm .....
bài này mình cũng ko làm đc
mà bạn hỏi giờ này ko ai trả lời đâu
tìm các số tự nhiên n trong khoang từ 100 đến 150 để phân số 3n+2 /7n-1 rút gọn dược
tìm các số tự nhiên n trong khoang từ 100 đến 150 để phân số 3n+2 /7n-1 rút gọn dược
Lời giải:
Gọi $d=ƯCLN(3n+2, 7n-1)$
$\Rightarrow 3n+2\vdots d; 7n-1\vdots d$
$\Rightarrow 7(3n+2)-3(7n-1)\vdots d$
$\Rightarrow 17\vdots d\Rightarrow d\in\left\{1; 17\right\}$
Để phân số rút gọn được thì $d=17$
Điều này xảy ra khi $3n+2\vdots 17$
$\Rightarrow 3n+2-17\vdots 17$
$\Rightarrow 3n-15\vdots 17$
$\Rightarrow 3(n-5)\vdots 17$
$\Rightarrow n-5\vdots 17$
$\Rightarrow n=17k+5$ với $k$ tự nhiên bất kỳ.
Vì $100\leq n\leq 150\Rightarrow 100\leq 17k+5\leq 150$
$\Rightarrow 5,6\leq k\leq 8,5$
Vì $k$ tự nhiên nên $k\in\left\{6;7;8\right\}$
$\Rightarrow n\in \left\{107; 124; 141\right\}$
Bài 2:
`c1:5/9 + ( (-4)/9 + 2 ) = (5/9 + (-4)/9 ) + 2 = 1/9 + 2 = 19/9 `
`c2:5/11 + ( (-6)/11 + 1 ) = ( 5/11+ (-6)/11 ) +1 = -1/11 + 1 = 10/11`
`c3:(-6)/13 + (1 + (-7)/13) = (-6/13+ (-7)/13 ) + 1 = -1 + 1 = 0`
`c4:(-7)/31+(99+ (-7)/31)=( (-7)/31+(-7)/31 )+99= -14/31+99= 3055/31 `
`d1: 1/2 + 1/3 + 1/4 + 1/6`
`= (6+4+3)/12 + 1/6`
`= 13/12 + 2/12`
`= 15/12 = 5/4`
`d2: 2/3 + (-3)/4 + 5/8 + (-1)/2 `
`= (16+(-18) +15)/24 +(-1)/2 `
`= 13/24 + (-1)/2 `
`= 13/24 + (-12)/24 `
`= 1/24`
`d3: 5/17 + (-9)/15 + (-2)/17 + 2/(-5)`
`= (5/17+ (-2)/17) + ((-9)/15 + (-6)/15) `
`= 3/17 + -15/15 `
`= 3/17 + (-1) `
`= -14/17`
`d4: -1/2 + 7/21 + (-2)/6 + (-6)/30 `
`= -1/2 + 1/3 + (-1/3) + (-1/5) `
`= (1/3 + (-1)/3 ) + ( (-5)/10 + (-2)/10 ) `
`= 0 + (-7)/10`
`= -7/10`
chứng minh rằng 1/1002 +1/1012+1/102.......+1/1992<1/99
Đặt \(S=\dfrac{1}{100^2}+\dfrac{1}{101^2}+...+\dfrac{1}{199^2}\)
\(S< \dfrac{1}{99.100}+\dfrac{1}{100.101}+...+\dfrac{1}{198.199}\)
\(S< \dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{100}-\dfrac{1}{101}+...+\dfrac{1}{198}-\dfrac{1}{199}\)
\(S< \dfrac{1}{99}-\dfrac{1}{199}< \dfrac{1}{99}\) (đpcm)
Giúp mình với
Tính nhanh:
\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)
(cảm ơn nhiều ạa)
\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)
\(3A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(3A=\dfrac{24}{49}\)
\(A=\dfrac{24}{49}:3\)
\(A=\dfrac{8}{49}\)
1) \(\dfrac{5}{13}\) +\(\dfrac{-5}{17}\) + \(\dfrac{-20}{41}\) + \(\dfrac{8}{13}\) +\(\dfrac{-21}{41}\)
2) \(\dfrac{1}{5}\) + \(\dfrac{-2}{9}\) + \(\dfrac{-7}{9}\) + \(\dfrac{4}{5}\) +\(\dfrac{16}{17}\)
3) \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) +\(\dfrac{2}{7.9}\) +....................+ \(\dfrac{2}{99.101}\)
1: =5/13+8/13-20/41-21/41-5/17
=1-1-5/17=-5/17
2: =1/5+4/5-2/9-7/9+16/17
=16/17+1-1=16/17
3: =1/3-1/5+1/5-1/7+...+1/99-1/101
=1/3-1/101
=98/303
j) \(\dfrac{-3}{4}\) + \(\dfrac{2}{7}\) + \(\dfrac{-1}{4}\) +\(\dfrac{3}{5}\) +\(\dfrac{5}{7}\)
k) \(\dfrac{-2}{17}\)+ \(\dfrac{15}{23}\) + \(\dfrac{-15}{17}\) + \(\dfrac{4}{19}\) +\(\dfrac{8}{23}\)
`@` `\text {Ans}`
`\downarrow`
`j)`
`-3/4 + 2/7 + (-1)/4 + 3/5 + 5/7`
`= (-3/4 - 1/4) + (2/7 + 5/7) + 3/5`
`= -1 + 1 + 3/5`
`= 3/5`
`k)`
`-2/17 + 15/23 + (-15)/17 + 4/19 + 8/23`
`= (-2/17 - 15/17) + (15/23 + 8/23) + 4/19`
`= -1 + 1 + 4/19`
`= 4/19`
$#KDN040510$
j: =-3/4-1/4+2/7+5/7+3/5
=-1+1+3/5
=3/5
k: =-2/17-15/17+15/23+8/23+4/19
=-1+1+4/19
=4/19
\(j)\dfrac{-3}{4}+\dfrac{2}{7}+\dfrac{-1}{4}+\dfrac{3}{5}+\dfrac{5}{7}\\ =\left(\dfrac{-3}{4}+\dfrac{-1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-4}{4}+\dfrac{7}{7}+\dfrac{3}{5}\\ =-1+1+\dfrac{3}{5}\\ =0+\dfrac{3}{5}\\ =\dfrac{3}{5}\\ k)\dfrac{-2}{17}+\dfrac{15}{23}+\dfrac{-15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\\ =\left(\dfrac{-2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\\ =\dfrac{-17}{17}+\dfrac{23}{23}+\dfrac{4}{19}\\ =-1+1+\dfrac{4}{9}\\ =0+\dfrac{4}{19}\\ =\dfrac{4}{19}\)