hời 2/3 + 2/3 =
a 4/3 b 3/4 c 1 d 4/2
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DC
3
9 tháng 8 2023
a) \(x+2\dfrac{3}{4}=5\dfrac{2}{3}\)
\(x+\dfrac{11}{4}=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}-\dfrac{11}{4}\)
\(x=\dfrac{35}{12}\)
b) \(x-1\dfrac{4}{5}=3\dfrac{2}{7}\)
\(x-\dfrac{9}{5}=\dfrac{23}{7}\)
\(x=\dfrac{23}{7}+\dfrac{9}{5}\)
\(x=\dfrac{178}{35}\)
c) \(x\times3\dfrac{1}{2}=4\dfrac{3}{4}\)
\(x\times\dfrac{7}{2}=\dfrac{19}{4}\)
\(x=\dfrac{19}{4}\div\dfrac{7}{2}\)
\(x=\dfrac{19}{14}\)
d) \(x\div2\dfrac{2}{3}=4\dfrac{1}{3}\)
\(x\div\dfrac{8}{3}=\dfrac{13}{3}\)
\(x=\dfrac{13}{3}\times\dfrac{8}{3}\)
\(x=\dfrac{104}{9}\)
ND
Nguyễn Đức Trí
VIP
9 tháng 8 2023
a) \(...=x+\dfrac{11}{3}=\dfrac{17}{3}\Rightarrow x=\dfrac{17}{3}-\dfrac{11}{3}=\dfrac{6}{3}=2\)
b) \(...\Rightarrow x-\dfrac{9}{5}=\dfrac{23}{7}\Rightarrow x=\dfrac{23}{7}+\dfrac{9}{5}=\dfrac{115}{35}+\dfrac{36}{35}=\dfrac{151}{35}\)
c) \(...\Rightarrow x.\dfrac{7}{2}=\dfrac{19}{4}\Rightarrow x=\dfrac{19}{4}:\dfrac{7}{2}\Rightarrow x=\dfrac{19}{4}.\dfrac{2}{7}=\dfrac{19}{14}\)
d) \(...\Rightarrow x:\dfrac{8}{3}=\dfrac{13}{3}\Rightarrow x=\dfrac{13}{3}.\dfrac{8}{3}=\dfrac{124}{9}\)
NV
26 tháng 6 2018
\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
\(b,B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{101}-1}{2}\)
\(c,C=4+4^2+4^3+...+4^n\)
\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3C=4^{n+1}-4\)
\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)
\(d,D=1+5+5^2+...+5^{2000}\)
\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(\Rightarrow D=\frac{5^{2001}-1}{4}\)
TC
21 tháng 3 2021
b)
B=1+3+3^2+3^3+..+3^100
=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)
=> 2B = 3^101 - 1
=> B =( 3^101 - 1) / 2
TH
0
A
A