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Bài 1:
\(a)\left(\dfrac{1}{3}:x-1\right)=\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{21}\\ \dfrac{1}{3}:x-1=\dfrac{7}{21}+\dfrac{3}{21}-\dfrac{1}{21}=\dfrac{9}{21}\\ \dfrac{1}{3}:x-1=\dfrac{3}{7}\\ \dfrac{1}{3}:x=\dfrac{3}{7}+1=\dfrac{10}{7}\\ x=\dfrac{1}{3}:\dfrac{10}{7}\\ x=\dfrac{7}{30}\\ b)\dfrac{1}{5}\cdot x-\dfrac{2}{13}=\dfrac{1}{2\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{5}{8\cdot13}\\ \dfrac{1}{5}\cdot x-\dfrac{2}{13}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{13}\\ \dfrac{1}{5}\cdot x-\dfrac{2}{13}=\dfrac{1}{2}-\dfrac{1}{13}=\dfrac{11}{26}\\\dfrac{1}{5}\cdot x=\dfrac{11}{26}+\dfrac{2}{13}=\dfrac{15}{26}\\ x=\dfrac{15}{26}:\dfrac{1}{5}=\dfrac{75}{26}\\c)\dfrac{13}{6} :\left(\dfrac{1}{2}+x\right)=\dfrac{1}{3}+\dfrac{3}{7}+\dfrac{1}{7\cdot2}+\dfrac{5}{2\cdot13}+\dfrac{3}{13\cdot4}\\ \dfrac{13}{6}:\left(\dfrac{1}{2}+x\right)=\dfrac{1}{3}+\left(\dfrac{3}{7}+\dfrac{1}{7\cdot2}\right)+\left(\dfrac{5}{2\cdot13}+\dfrac{3}{13\cdot4}\right)\\ \dfrac{13}{6}:\left(\dfrac{1}{2}+x\right)=\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{4}\\ \dfrac{13}{6}:\left(\dfrac{1}{2}+x\right)=\dfrac{13}{12}\\ \dfrac{1}{2}+x=\dfrac{13}{6}:\dfrac{13}{12}=2\\ x=2-\dfrac{1}{2}\\ x=\dfrac{3}{2}\)
19:
a: 71-(x+33)=26
=>x+33=71-26=45
=>x=45-33=12
b: \(\left(x-73\right)\cdot10^2-26=74\)
=>\(100\left(x-73\right)=26+74=100\)
=>x-73=1
=>x=73+1=74
c: \(\left(x+1\right)^3-4=60\)
=>\(\left(x+1\right)^3=4+60=64=4^3\)
=>x+1=4
=>x=3
17:
a: \(3\cdot5^2+15\cdot2^2-26:2\)
\(=3\cdot25+15\cdot4-13\)
=75+60-13
=135-13=122
b: \(37\cdot39+62\cdot21-11\cdot39-21\cdot36\)
\(=39\left(37-11\right)-21\left(62-36\right)\)
\(=39\cdot26-21\cdot26=26\cdot18=468\)
c: \(3^2\cdot5+2^2\cdot10-3^4:3\)
\(=9\cdot5+4\cdot10-3^3\)
=45+40-27
=45+13=58
d: Sửa đề: \(99-96+93-90+...-6+3\)
=(99-96)+(93-90)+...+(9-6)+3
=3+3+...+3
=3x16+3=48+3=51
còn anh/chị nào lớp 7 trở lên biết làm những bài này thì cíu elm vớiiii :((
1. They are speaking English now.
2.The sun is shining
3. It is raining right at the moment.
4. The wind is blowing right now.
5. Is she decorating the room now?
6. He is looking at the Christmas tree.
7. Is Mr. picke doing his homework?
8. She and her friend are swimming in the river.
9. We are watching television right now?
10. Are they playing in the garden now?
1. They/ speak/be/English/now.
They are speaking English now
2. Shine/ the sun/be.
The sun is shining
3. Be/it/rain/ right at the moment.
It is raining right at the moment
4. Blow/ the wind/ right now/be.
The wind is blowing right now
5. Decorate/she/be/now/ the/room?
Is she decorating the room now?
\(1,a)\dfrac{15}{12}-\dfrac{-1}{4}\\ =\dfrac{15}{12}+\dfrac{1}{2}\\ =\dfrac{15}{12}+\dfrac{6}{12}\\ =\dfrac{21}{12}=\dfrac{7}{4}\\ b)-\dfrac{5}{12}+0,75\\ =-\dfrac{5}{12}+\dfrac{3}{4}\\ =\dfrac{-5}{12}+\dfrac{9}{12}\\ =\dfrac{4}{12}=\dfrac{1}{3}\\ c)\dfrac{15}{12}+\dfrac{5}{13}-\left(\dfrac{3}{12}+\dfrac{18}{13}\right)\\ =\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}\\ =\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\\ =\dfrac{12}{12}-\dfrac{13}{13}\\ =1-1=0\)
2: a: \(-\dfrac{16}{42}-\dfrac{5}{8}=\dfrac{-64}{168}-\dfrac{105}{168}=\dfrac{-169}{168}\)
b: \(3,5-\left(-\dfrac{2}{7}\right)=3,5+\dfrac{2}{7}=\dfrac{7}{2}+\dfrac{2}{7}=\dfrac{7^2+2^2}{14}=\dfrac{53}{14}\)
c: \(\left(-\dfrac{1}{2}+\dfrac{3}{4}\right)-\left(-\dfrac{4}{5}+\dfrac{5}{6}\right)\)
\(=\dfrac{-1}{2}+\dfrac{3}{4}+\dfrac{4}{5}-\dfrac{5}{6}\)
\(=\dfrac{-30}{60}+\dfrac{45}{60}+\dfrac{48}{60}-\dfrac{50}{60}\)
\(=\dfrac{15}{60}-\dfrac{2}{60}=\dfrac{13}{60}\)
3:
a: \(\dfrac{2}{21}-\dfrac{-1}{28}=\dfrac{2}{21}+\dfrac{1}{28}=\dfrac{8}{84}+\dfrac{3}{84}=\dfrac{11}{84}\)
b: \(-4.75-1\dfrac{7}{12}=-\dfrac{57}{12}-\dfrac{19}{12}=-\dfrac{76}{12}=-\dfrac{19}{3}\)
c: \(-\left(\dfrac{3}{5}+\dfrac{5}{4}\right)-\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\)
\(=-\dfrac{3}{5}-\dfrac{5}{4}+\dfrac{3}{4}-\dfrac{2}{5}\)
\(=-1-\dfrac{2}{4}=-\dfrac{3}{2}\)
4:
a: \(-\dfrac{2}{33}+\dfrac{5}{55}=\dfrac{-10}{165}+\dfrac{15}{165}=\dfrac{5}{165}=\dfrac{1}{33}\)
b: \(0,4+\left(-2\dfrac{4}{5}\right)=0,4-2,8=-2,4\)
c: \(-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\)
\(=\dfrac{-3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\)
\(=-\dfrac{3}{7}-\dfrac{4}{7}=-\dfrac{7}{7}=-1\)
1 walks
2 are learning
3 is going
4 feel
5 are studying
6 have
7 does
8 likes
9 What time do you have lunch everyday?
10 doesn't have
Gọi số học sinh đi tham quan là \(a\)
Điều kiện: \(a\inℕ^∗;700\le a\le1200\)
Ta có:
+) Nếu xếp 30 em hay 45 em vào 1 xe thì đều thiếu 5 em
⇒\(a\) chia \(30\) hay \(45\) thiếu \(5\)
\(\Rightarrow a+5⋮30;45\)
\(\Rightarrow a+5\in BC\left(30;45\right)=\left\{0,90,180,270,360,450,540,630,720,810,900,990,1080,1170,1260,...\right\}\)
Mà \(700\le a\le1200\) nên \(705\le a+5\le1205\) suy ra:
\(a\in\left\{720,810,900,990,1080,1170\right\}\)
+) Nếu xếp 43 em vào một xe thì vừa đủ
\(\Rightarrow a⋮43\)
Do đó: \(a=1075\) (thỏa mãn điều kiện)
Vậy...
Gọi tổng số h/s là A
A:30 thiếu 5 , chia 45 cũng thiếu 5 ≠Ta có :
A+5 ∈ BCNN(45,30)700≤A≤1200
30=2.3.5
45=2.3.3.5=2.32.5
BCNN(30,45)=2.95=90
BC(30,45)={0,90,180,270,360,450,540,630,720,810,900,990,1080,1170} mà 700≤A≤1200 nên loại các số 0,90,180,270,360,450,540,630.
Nếu A là 1 trong các số trên thì phải trừ đi 5 , A ∈={715,805,895,985,1075,1165}
Vì A⋮43 nên A sẽ bằng 1075 , vậy chuyến đi đó có 1075 h/s lớp 6
Đáp số 1075 h/s
Gọi số đó có dạng \(\overline{ab}\)
Khi thêm số 0 vào giữa thì ta có số mới là: \(\overline{a0b}=100a+b\)
Mà số mới gấp 7 lần số cũ nên ta có:
\(\overline{a0b}=7\overline{ab}\\ 100a+b=7\left(10a+b\right)\\ 100a+b=70a+7b\\ 100a-70a=7b-b\\ a\left(100-70\right)=b\left(7-1\right)\\ 30a=6b\\ \dfrac{a}{b}=\dfrac{6}{30}=\dfrac{1}{5}\)
`=> a=1;b=5`
Vậy sso cần tìm là 15
câu b đề bài cho B < \(\dfrac{1}{3}\) thì cần gì chứng minh nữa em.
Đáng lẽ phải là: Cho B = \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) +...-\(\dfrac{1}{64}\)
Chứng minh B < \(\dfrac{1}{3}\)
Gọi số cần tìm có dạng là \(\overline{ab}\)
Nếu viết thêm chữ số 0 vào giữa hai chữ số thì ta được số mới gấp 6 lần số cũ nên \(\overline{a0b}=6\cdot\overline{ab}\)
=>\(100a+b=6\cdot\left(10a+b\right)\)
=>100a+b=60a+6b
=>40a=5b
=>8a=b
=>b=8; a=1
Vậy: Số cần tìm là 18