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\(\frac{9}{2}-\frac{8}{3}:\frac{2}{3}+\frac{1}{2}=\frac{9}{2}-4+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1\)
\(\frac{12}{5}:\frac{4}{7}-\frac{1}{2}=\frac{21}{5}-\frac{1}{2}=\frac{37}{10}\)
A) 9/2 - 8/3 : 2/3 + 1/2 = 9/2 - 8/3 x 3/2 + 1/2
= 9/2 - 4 + 1/2
= 1/2 + 1/2
= 1.
B) 12/5 : 4/7 - 1/2 = 12/5 x 7/4 - 1/2
= 21/5 -1/2
= 21/10.
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a) Với m = 0, giá trị biểu thức 12 : (3 – m) là:
12 : (3 – 0) = 12 : 3 = 4
Với m = 1, giá trị biểu thức 12 : (3 – m) là:
12 : (3 – 1 ) = 12 : 2 = 6
Với m = 2, giá trị biểu thức 12 : (3 – m) là:
12 : (3 – 2) = 12 : 1 = 12
b) Vì 4 < 6 < 12 nên trong ba giá trị tìm được ở câu a, với m = 2 thì biểu thức 12 : (3 – m) có giá trị lớn nhất.
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ta có:
1/2+1/6+...+1/9900
=1/1.2+1/2.3...+1/99.100
=1-1/2+1/2-1/3+1/3-...+1/99-1/100
=1-1/100
=99/100
\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{9900}\)
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\cdot\cdot\cdot+\frac{1}{99\times100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
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1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân S với 3 ta có:
S x 3 = 3 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Vậy:
S x 3 - S = 3 - 1/243
2 S = 2186 / 729
S = 2186 / 729 : 2
S = 1093/729
S+3 =1093/729 + 3
=3280/729
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a×1/2+a×1/6+a×1/12
vơi a =8
\(\Rightarrow\)8×1/2+8×1/6+8×1/12=8×(1/2×1/6+1/12)=8×3/4=6
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\(\dfrac{1}{2}\times\dfrac{3}{4}:\dfrac{6}{5}=\dfrac{3}{8}:\dfrac{6}{5}=\dfrac{5}{16}\)
\(\dfrac{2}{3}+\dfrac{1}{6}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{14}{21}+\dfrac{6}{21}=\dfrac{20}{21}\)
\(8\times\dfrac{3}{5}:\dfrac{12}{5}=\dfrac{24}{5}:\dfrac{12}{5}=2\)
\(\dfrac{4}{9}+\dfrac{3}{4}-\dfrac{1}{3}=\dfrac{16}{36}+\dfrac{27}{36}-\dfrac{12}{36}=\dfrac{31}{36}\)
`12/3-1/2:2`
`=4-1/2xx1/2`
`=4-1/4`
`=16/4-1/4`
`=15/4`
12/3-1/2:2
=12/3-1/4
=15/4