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bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
Bài 1: Thực hiện các phép tính sau:
\(a)\)Chưa rỏ đề
\(b)\)\(5025\div5-25\div5\)
\(=\)\(1005-5\)
\(=\)\(1000\)
\(c)\)\(218-180\div2\div9\)
\(=\)\(218-10\)
\(=\)\(208\)
\(d)\)\(\left(328-8\right)\div32\)
\(=\)\(320\div32\)
\(=\)\(10\)
Bài 1:
a) ( Tôi không nhìn rõ đầu bài )
b) 5025 : 5 - 25 : 5
= ( 5025 - 25 ) : 5
= 5000 : 5
= 1000
c) 218 - 180 : 2 : 9
= 218 - 180 : ( 2 . 9 )
= 218 - 180 : 18
= 218 - 10
= 208
d) ( 328 - 8 ) : 32
= 320 : 32
= 10
4:
a: =4/15-2,9+11/15=1-2,9=-1,9
b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)
c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)
d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)
e: =-5/12-7/12-3,7-6,3=-1-10=-11
f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10
Mk thấy bài 1 và 2 dễ nên bạn tự làm nha
3
+)Ta có n-2 \(⋮\)n-2
=>2.(n-2)\(⋮\)n-2
=>2n-4\(⋮\)n-2(1)
+)Theo bài ta có:2n+1\(⋮\)n-2(2)
+)Từ (1) và (2)
=>(2n+1)-(2n-4)\(⋮\)n-2
=>2n+1-2n+4\(⋮\)n-2
=>5\(⋮\)n-2
=>n-2\(\in\)Ư(5)={\(\pm\)1;\(\pm\)5}
+)Ta có bảng:
| n-2 | -1 | 1 | -5 | 5 |
| n | 1\(\in\)Z | 3\(\in\)Z | -3\(\in\)Z | 7\(\in\)Z |
Vậy n\(\in\){1;3;-3;7}
Chúc bn học tốt
a. 5.(–8).( –2).(–3) b. 4.(–5)2 + 2.(–5) – 20
=(-5).8.(-2).(-3) ={(-5).2} {4+1}-20
=(-5)(-2)(-3).8 =(-10).5-20=-50-20=-70
=10.(-24)=-240
\(\frac{6}{11}\cdot\frac{3}{14}+\frac{-5}{16}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-3}{16}\)
\(=\frac{6}{11}\cdot\frac{3}{14}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-5}{16}+\frac{-3}{16}\)
\(=\frac{3}{14}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{-5}{16}+\frac{-3}{16}\right)\)
\(=\frac{3}{14}\cdot\frac{6+5}{11}+\frac{-5+\left(-3\right)}{16}\)
\(=\frac{3}{14}\cdot\frac{11}{11}+\frac{-8}{16}\)
\(=\frac{3}{14}\cdot1+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-7}{14}\)
\(=\frac{3+\left(-7\right)}{14}\)\(=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-5}{6}+\left(7x-\frac{1}{2}\right)\cdot\frac{2}{9}=-1\frac{1}{3}\)
\(\frac{-5}{6}+\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}+\frac{5}{6}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{1}{2}\)
\(\frac{14}{9}\cdot x=-\frac{1}{2}+\frac{1}{9}\)
\(\frac{14}{9}\cdot x=-\frac{7}{18}\)
\(x=-\frac{7}{18}:\frac{14}{9}\)
\(x=-\frac{1}{4}\)
a) 1-3+5-7+9-12+15-18
= (1+ 9) - (3+7) + (5+15) - (12+18)
= 10 - 10 + 20 - 30 = -10.
b) (-2)+5-7+9-11+13-15+19-21
= 5 - 11 - 2 -2 = 5 - (11+2+2)
= 5 - 15 = -10
Bài 1:
a) Ta có: 1-3+5-7+9-12+15-18
=(1+9)-(3+7)+(5+15)-(12+18)
=10-10+20-30
=-10
b) Ta có: \(\left(-2\right)+5-7+9-11+13-15+19-21\)
\(=3+2+2+4-21\)
\(=5+6-21\)
=11-21=-10
Bài 1 :
a ) ( 2x + 1 )3 = 9.81
( 2x + 1 )3 = 729
( 2x + 1 )3 = 93
=> 2x + 1 = 9
2x = 9-1 = 8
x = 8 : 2 = 4
Vậy x = 4
b ) ( 7x - 11 )3 = 25 . 52 + 200
( 7x - 11 )3 = 32 . 25 + 200
( 7x - 11 )3 = 1000
( 7x - 11 )3 = 103
=> 7x - 11 = 10
7x = 10 + 11 = 21
x = 21 : 7 = 3
Vậy x = 3
Bài 2 :
a ) 50 - [ 30.( 6-2 )2 ]
= 50 - [ 30. 42 ]
= 50 - [ 30. 16 ]
= 50 - 480
= -430
b ) Tương tự
Bài 3 :
Câu hỏi của Phạm Ngọc Thạch - Toán lớp 6 - Học toán với OnlineMath
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