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Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!
a: =91/105+60/105-101/105
=50/105=10/21
c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)
d: =2-2/9
=18/9-2/9
=16/9
e: =24/36-9/36+8/36
=23/36
g: =5/2+1/2
=3
\(\left(\dfrac{6:\dfrac{3}{5}-\dfrac{17}{16}.\dfrac{6}{7}}{\dfrac{21}{5}.\dfrac{10}{11}+\dfrac{57}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{15}\right).\dfrac{12}{49}}{\dfrac{10}{3}+\dfrac{2}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{\dfrac{509}{56}}{9}-\dfrac{\dfrac{7}{12}.\dfrac{12}{49}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{\dfrac{1}{7}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{9}{224}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\dfrac{1955}{2016}.x=\dfrac{215}{96}\)
\(\Rightarrow x=\dfrac{215}{96}:\dfrac{1955}{2016}\)
\(\Rightarrow x=\dfrac{903}{391}\)
`[ 6 : 3/5 - 17/16 . 6/7 : 21/5 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 : 10/3 + 2/9 ] . x = 215/96`
`=>[ 6 . 5/3 - 17/16 . 6/7 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=>[10- 51/56 . 6/7 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 153/196 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 255/1372 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 1275/7546 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> (10- 1275/7546 + 57/11 - 7/12. 12/49 . 3/10 + 2/9 ) . x = 215/96`
`=> ( 10- 1275/7546 + 57/11 -343/600 . 3/10 + 2/9 ) . x = 215/96`
`=> ( 10- 1275/7546 + 57/11 -343/2000 + 2/9 ) . x = 215/96`
`=>15,06357671 . x= 215/96`
`=> x= 215/96: 15,06357671`
`=>x= 0,1486754027`
Đề có phải như vậy không nhỉ ?
Bài 1 :
\(A=1+2+3+4+5+6+7+8+9\)
\(=\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5\)
\(=10+10+10+10+5\)
\(=45\)
Bài 2 bạn tự làm nha
Bài 2 : Tìm x :
a) ( x - 8 )5 = ( x - 8 )9
=> x - 8 thuộc { 0 ; 1 } vì :
Nếu x - 8 = 0 thì ( x - 8 )5 = 05 = 0
Nếu x - 8 = 0 thì ( x - 8 )9 = 09 = 0
Vậy x - 8 = 0 là thỏa mãn điều kiện.
Nếu x - 8 = 1 thì ( x - 8 )5 = 15 = 1
Nếu x - 8 = 1 thì ( x - 8 )9 = 19 = 1
Vậy x - 8 = 1 là thỏa mãn điều kiện.
=> x - 8 thuộc { 0 ; 1 }
=> x thuộc { 8 ; 9 }