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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
Bài 1:
1: =-5/24+16/27+3/4
=-5/24+18/24+16/27
=13/24+16/27
=117/216+128/216=245/216
2: =-1/3+1/3+6/7=6/7
3: \(=\dfrac{1}{2}-\dfrac{7}{12}+\dfrac{1}{2}=1-\dfrac{7}{12}=\dfrac{5}{12}\)
4: \(=-\dfrac{5}{8}+\dfrac{14}{25}-\dfrac{6}{10}=\dfrac{-125+112-120}{200}=\dfrac{-133}{200}\)
\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
Chúc bạn học tốt ~
1,
a,-3/5
b,-1/2
c,19/39
d,1/4
e,-39/40
f,-59/56
2,
a,=
b,<
c,>
d,<
k cho mình nha
1) x - 2 = -6
x = -6 + 2
x = -4
2) -5 . x - ( -3 ) =13
-5 . x = 13 + ( -3 )
-5 . x = 10
x = 10 : ( -5 )
x = -2
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= \(\frac{1}{5}\)-\(\frac{1}{12}\)=\(\frac{7}{60}\)