Rút gọn :
D = 21 + 24 + 27 + 210 + ............ + 2100
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a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)
\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)
\(=\dfrac{2591}{2^6\cdot3^5}\)
\(a,\dfrac{8}{24}-\dfrac{5}{25}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
\(b,\dfrac{18}{21}-\dfrac{12}{27}=\dfrac{6}{7}-\dfrac{4}{9}=\dfrac{26}{63}\)
a) \(\dfrac{8}{24}-\dfrac{5}{25}=\dfrac{8:8}{24:8}-\dfrac{5:5}{25:5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
b) \(\dfrac{18}{21}-\dfrac{12}{27}=\dfrac{18:3}{21:3}-\dfrac{12:3}{27:3}=\dfrac{6}{7}-\dfrac{4}{9}=\dfrac{54}{63}-\dfrac{28}{63}=\dfrac{26}{63}\)
45/27=5/3
25/50=1/2
12/21=4/7
24/56=3/7
22/121=2/11
196/28=7
26/169=2/13
221/187=13/11
1/
Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.
Số số hạng: $(101-1):4+1=26$
$A=(101+1)\times 26:2=1326$
2/
$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$
$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$
$=(1+2+2^2)(1+2^3+2^6+2^9)$
$=7(1+2^3+2^6+2^9)\vdots 7$
A = 1 + 3 + 32 + 33 + ... + 3100
3A = 3 + 32 + 33 +34+ .... + 3101
3A - A = (3 + 32 + 34 + ... + 3101) - (1 + 3 + 32 + 33 + ... + 3100)
2A = 3 + 32 + 34 + ... + 3101 - 1 - 3 - 32 - 33 - ... - 3100
2A = (3 - 3) + (32 - 32) + ... + (3100 - 3100) + (3101 - 1)
2A = 3101 - 1
A = \(\dfrac{3^{101}-1}{2}\)
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
\(\frac{3}{5}-\frac{5}{35}=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)
\(\frac{18}{27}-\frac{2}{6}=\frac{1}{3}\)
\(\frac{15}{25}-\frac{3}{21}=\frac{16}{35}\)
\(\frac{24}{36}-\frac{6}{12}=\frac{1}{6}\)
tích đúng cho mk nhé
35−535=2135−535=163535−535=2135−535=1635
1827−26=131827−26=13
1525−321=16351525−321=1635
2436−612=16
Câu a cả trên lẫn dưới chia cho 18 ra 4/3
Câu b cả trên lẫn dưới chia cho 6 ra 4/5
Câu c cả trên lẫn dưới chia cho 20 ra 3/4
Câu d cả trên lẫn dưới chia cho 12 ra 5/3
Câu e cả trên lẫn dưới chia cho 35 ra 1/6
Một k nha
\(D=2^1+2^4+2^7+...+2^{100}\)
\(2^3D=8D=2^4+2^7+2^{10}+...+2^{103}\)
\(8D-D=6D=2^{103}-2\Rightarrow D=\frac{2^{103}-2}{6}\)
Nhầm khúc cuối: \(8D-D=7D=2^{103}-2\Rightarrow D=\frac{2^{103}-2}{7}\)