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a) \(1+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(=\frac{16}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\)
\(=\frac{23}{16}\)
b) \(2-\frac{1}{8}-\frac{1}{12}-\frac{1}{16}\)
\(=\frac{96}{48}-\frac{6}{48}-\frac{4}{48}-\frac{3}{48}\)
\(=\frac{83}{48}\)
c) \(\frac{4}{99}\cdot\frac{18}{5}\div\frac{12}{11}+\frac{3}{5}\)
\(=\frac{4\cdot18\cdot11}{99\cdot5\cdot12}+\frac{3}{5}\)
\(=\frac{4\cdot9\cdot2\cdot11}{9\cdot11\cdot5\cdot4\cdot3}+\frac{3\cdot3}{3\cdot5}\)
\(=\frac{2}{15}+\frac{9}{15}=\frac{11}{15}\)
d) \(\left(1-\frac{3}{4}\right)\left(1+\frac{1}{3}\right)\div\left(1-\frac{1}{3}\right)\)
\(=\frac{1}{4}\cdot\frac{4}{3}\div\frac{2}{3}\)
\(=\frac{1\cdot4\cdot3}{4\cdot3\cdot2}=\frac{1}{2}\)
Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)
=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)
=> x = 9
Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)
=> \(\frac{15}{16}:x=\frac{11}{12}\)
=> \(x=\frac{45}{44}\)
Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)
=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
=> \(\frac{1}{x+1}=\frac{1}{800}\)
=> x = 799
Bài 2 :
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)
Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(=1-\frac{1}{12}=\frac{11}{12}\) (2)
Thay (1) và (2) vào biểu thức (*) ta được :
\(\frac{15}{16}:x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)
\(\Leftrightarrow x=\frac{45}{44}\)
Vậy : \(x=\frac{45}{44}\)
a) x + 2/3=9/11
<=> x=9/11-2/3= 5/33
b) x - 3/10=4/15
<=> x=4/15+3/10=17/30
c) X x 1/7=5/6
<=> x=5/6:1/7=35/6
d) x : 3/5=1/6
<=> x=1/6x3/5=1/10
\(a,x+\frac{2}{3}=\frac{9}{11}\)
\(x=\frac{9}{11}-\frac{2}{3}\)
\(x=\frac{5}{33}\)
\(b,x-\frac{3}{10}=\frac{4}{15}\)
\(x=\frac{4}{15}+\frac{3}{10}\)
\(x=\frac{17}{30}\)
\(c,x\cdot\frac{1}{7}=\frac{5}{6}\)
\(x=\frac{5}{6}:\frac{1}{7}\)
\(x=\frac{35}{6}\)
\(d,x:\frac{3}{5}=\frac{1}{6}\)
\(x=\frac{1}{6}\cdot\frac{3}{5}\)
\(x=\frac{1}{10}\)
. là nhân nha
a) 12/17 + 21/25 = 657/425
b) 4 - 205/811 = 3039/811
c) 12 . 21/25 = 252/25
d) 4 : 205/811 = 205/3244
a) 12/17 + 21/25 = 657/425
b) 4 - 205/811 = 3039/811
c) 12 . 21/25 = 252/25
d) 4 : 205/811 = 205/3244
a, \(1\frac{2}{7}+6\frac{5}{6}=\frac{9}{7}+\frac{41}{6}=\frac{54}{42}+\frac{287}{42}=\frac{341}{42}\)
a. \(\frac{5}{12}:x=\left(\frac{5}{6}-\frac{1}{2}\right)\times\frac{5}{4}\)
\(=>\)\(\frac{5}{12}:x=\frac{1}{3}\times\frac{5}{4}=\frac{5}{12}\)
\(=>x=\frac{5}{12}:\frac{5}{12}=1\)
b.\(x:0,15=\left(2,8+0,38\right):0,75\)
\(=>x:0,15=3,18:0,75=4,24\)
\(=>x=4,24\times0,15=0,636\)
a. Ta tính trước số bị chia: 1 + 4 + 7 + …… + 100
Dãy số gồm có: (100 – 1) : 3 + 1 = 34 (số hạng)
Ta thấy: 1 + 100 = 4 + 97 = 101 = …..
Do đó số bị chia là: 101 x 34 : 2 = 1717
Ta có: 1717 : a = 17
a = 1717 : 17
a = 101
vậy a = 101.
b.
x - 1 2 × 5 3 = 7 4 - 1 2 x - 1 2 × 5 3 = 5 4 x - 1 2 = 5 4 : 5 3 x - 1 2 = 3 4 x = 3 4 + 1 2 x = 5 4
c. 2000 2001 v à 2001 2002
Ta có: 1 - 2000 2001 = 1 2001
1 - 2001 2002 = 1 2002
Vì 1 2001 > 1 2002 nên 2000 2001 < 2001 2002
Bài 1:
\(S=1+3+5+7+...+297+299\)
Tổng trên là tổng các số hạng cách đều, số hạng sau hơn số hạng trước \(2\)đơn vị.
Số số hạng của tổng trên là: \(\left(299-1\right)\div2+1=150\)(số hạng)
Giá trị của tổng trên là: \(\left(299+1\right)\times150\div2=22500\)
Bài 2:
\(100-7\times\left(x-5\right)=58\)
\(\Leftrightarrow7\times\left(x-5\right)=100-58\)
\(\Leftrightarrow7\times\left(x-5\right)=42\)
\(\Leftrightarrow x-5=42\div7\)
\(\Leftrightarrow x-5=6\)
\(\Leftrightarrow x=6+5\)
\(\Leftrightarrow x=11\)
mk cần gấp ak