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\(1+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{5}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{11}{8}+\dfrac{1}{16}\)
\(=\dfrac{23}{16}\)
______
\(2-\dfrac{1}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{15}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{43}{24}-\dfrac{1}{16}\)
\(=\dfrac{83}{48}\)
_________
\(\dfrac{4}{99}\times\dfrac{18}{5}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}\times\dfrac{11}{12}+\dfrac{3}{5}\)
\(=\dfrac{2}{15}+\dfrac{3}{5}\)
\(=\dfrac{11}{15}\)
__________
\(\left(1-\dfrac{3}{4}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{4}\times\dfrac{4}{3}\times\dfrac{2}{3}\)
\(=\dfrac{4\times2}{4\times3\times3}\)
\(=\dfrac{2}{3\times3}\)
\(=\dfrac{2}{9}\)
a) 1 + 1/4 + 1/8 + 1/16
= 16/16 + 4/16 + 2/16 + 1/16
= 23/16
b) 2 - 1/8 - 1/12 - 1/16
= 96/48 - 6/46 - 4/48 - 3/48
= 83/48
c) 4/99 × 18/5 : 12/11 + 3/5
= 8/55 : 12/11 + 3/5
= 2/15 + 3/5
= 2/15 + 9/15
= 11/15
d) (1 - 3/4) × (1 + 1/3) : (1 - 1/3)
= 1/4 × 4/3 : 2/3
= 1/3 : 2/3
= 2
a, 1. \(\dfrac{2}{17}\)
2. \(22\)
3. 4
b,
1. \(\dfrac{8}{5}\)
2. \(\dfrac{176}{5}\)
3. \(\dfrac{28}{5}\)
4. \(\dfrac{3}{10}\)
5. \(\dfrac{163}{5}\)
6. \(\dfrac{9}{20}\)
a) \(\dfrac{12}{17}:6=\dfrac{12}{17}\cdot\dfrac{1}{6}=\dfrac{2}{17}\)
5/8x16/7x21/25
=1/1x2/1x3/5
=6/5
11/12:22/30x6/15
=11/12x30/22x6/15
=1/2x1/2x1/2
=1/8
5/8x16/7x21/25
=1/1x2/1x3/5
=6/5
11/12:22/30x6/15
=11/12x30/22x6/15
=1/2x1/2x1/2
=1/8
1+(2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17)x0,000001
=1+(2+(3+17x10)+10)x0,000001
=1+(2+200+10)x0,000001
=1+212x0,000001
=1+0,000212
=1,000212
1+2-3-4+5+6-...+301+302
đặt A=(1+2)+(5+6)+...+(301+302)=3+11+19+...+603
A=[(603-3):8+1]x(603+3)=75x606=45450
đặt B=-3-4-7-8-11-13-...-299-300
B=(-3-4)+(-7-8)+(-11-12)+..+(-299-300)
B=-(7+15+23+...+599)
B=-[(599-7):8+1]x(599+7)=45450
ta có:
1+2-3-4+5+6-7-8+...+301+302=A-B=45450-45450=0
8/12 < 16/... < 8/11
16/24 < 16/... < 16/22
16/24 < 16/23 < 16/22
Vậy số cần tìm là 23