1) Tính:
B=1+2+2^2+2^3+...+2^2008/1-2^2009
B=-9/10.5/14+1/10.(-9/2)+1/7.9-9/10)
S=3+3/2+3/2^2+...+3/2^9
A=1/30+1/42+1/56+1/72+1/90+1/110+1/132
2) Diền vào chỗ trống thich hợp ?giải thích?
-1/2<.../7<.../14<.../7<.../14<-1/7
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Gọi tử số của B là a và mẫu là b
\(a=1+2+2^2+2^3+...+2^{2008}\)
\(2a=2+2^2+2^3+...+2^{2009}\)
\(2a-a=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\)
\(a=2^{2009}-1\)
\(a=\frac{2^{2009}-1}{1-2^{2009}}\)
\(a=1\)
$2a-a=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)$2a−a=(2+22+23+...+22009)−(1+2+22+...+22008)
$a=\left(2-2\right)+\left(2^2-2^2\right)+...+\left(2^{2008}-2^{2008}\right)+2^{2009}-1$a=(2−2)+(22−22)+...+(22008−22008)+22009−1
$a=0+0+0+2^{2009}-1$a=0+0+0+22009−1
$a=2^{2009}-1$a=22009−1
$B=\frac{2^{2009}-1}{1-2^{2009}}$B=22009−11−22009
B= -1
a; [6.(- \(\dfrac{1}{3}\))3 - 3.(- \(\dfrac{1}{3}\) + 1)] - ( - \(\dfrac{1}{3}\) - 1)
= [6. \(\dfrac{-1}{3^3}\) - 3.\(\dfrac{2}{3}\)] - ( - \(\dfrac{1}{3}\) - \(\dfrac{3}{3}\))
= [\(\dfrac{-2}{9}\) - 2] + \(\dfrac{4}{3}\)
= [\(\dfrac{-2}{9}\) - \(\dfrac{18}{9}\)] + \(\dfrac{12}{9}\)
= - \(\dfrac{20}{9}\) + \(\dfrac{12}{9}\)
= \(\dfrac{-8}{9}\)
b; (63 + 3.62 + 33): 13
= (216 + 3.36 + 27) : 13
= (216 + 108 + 27): 13
= (324 + 27): 13
= 351 : 13
= 27
a) \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
b) \(=\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{10}{30}-\dfrac{3}{30}=\dfrac{7}{30}\)
a , \(\frac{7}{8}:\frac{1}{6}+\frac{7}{8}.\frac{-7}{18}\)
= \(\frac{21}{4}+\frac{-49}{144}=\frac{707}{144}\)
b, -1 : (-5) + \(\frac{1}{15}-\frac{-1}{15}\)
= \(\frac{1}{5}+0=\frac{1}{5}\)
c, \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{10-9}{10.9}-\frac{9-8}{9.8}-\frac{8-7}{8.7}-\frac{7-6}{7.6}-\frac{6-5}{6.5}-\frac{5-4}{5.4}-\frac{4-3}{4.3}-\frac{3-2}{3.2}.\frac{2-1}{2.1}\)
= \(\frac{9}{10}-1-\frac{1}{10}-1-\frac{1}{9}-1-\frac{1}{8}-1-\frac{1}{7}-1-\frac{1}{6}-1-\frac{1}{5}-1-\frac{1}{4}-1-\frac{1}{3}-1-\frac{1}{2}\)
= \(\frac{9}{10}-\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}\right)\)
= \(\frac{9}{10}-9-1,928=\frac{9}{10}-7,071=-6.171\)
\(1\frac{1}{2}+2\frac{1}{6}+3\frac{1}{12}+4\frac{2}{20}+5\frac{1}{30}+6\frac{1}{42}+7\frac{1}{56}+8\frac{1}{72}+9\frac{1}{90}+\frac{1}{10}\)\(=\frac{3}{2}+\frac{13}{6}+\frac{37}{12}+\frac{81}{20}+\frac{151}{30}+\frac{253}{42}+\frac{393}{56}+\frac{577}{72}+\frac{811}{90}+\frac{1}{10}=46\)
k nha
๖ۣۜH๖ۣۜU๖ۣۜY ๖ۣۜR๖ۣۜI๖ۣۜO
Đầu tiên , cộng các phần nguyên lại với nhau , ta có :
( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( 12 +16 +112 +120 +130 +142 +156 +172 +190 +110 )
= 45 + (16 +130 )+12 +112 +120 +142 +156 +172 +190 +110
sau khi cộng trong ngoặc , ta được 6 / 30 , rút gọn tối giản còn 1 / 5
= 45 + (15 +120 )+12 +112 +142 +156 +172 +190 +110
sau khi cộng trong ngoặc và rút gọn tối giản , ta được 1 / 4
= 45 + (14 +12 )+112 +142 +156 +172 +190 +110
sau khi cộng trong ngoặc rồi rút gọn , ta được 3 / 4
= 45 + (34 +112 )+142 +156 +172 +190 +110
rút gọn lại ta được 5 / 6
= 45 + (56 +142 )+156 +172 +190 +110
rút gọn tối giản ra 6 / 7
= 45 + (67 +156 )+172 +190 +110
sau khi tính trong ngoặc rút gọn được 7 / 8
= 45 + (78 +172 )+190 +110
tính trong ngoặc rồi rút gọn ra 8 / 9
= 45 + (89 +190 )+110
cũng rút gọn tiếp ta được 9 / 10
= 45 + (910 +110 )
= 45 + 1
= 46
1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)
\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)
\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)
A=\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)+\(\frac{1}{90}\)+\(\frac{1}{110}\)+\(\frac{1}{132}\)
A=\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{12}\)=\(\frac{7}{60}\)