Câu 1
Cho P= 1+2+ 2^2 + 2^3+2^4 + 2^5 + 2^6 + 2^7
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Câu 1:
a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$
b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)
c.
\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)
d.
\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)
Câu 2:
a. $x\times \frac{2}{7}=\frac{6}{11}$
$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$
b. $x: \frac{3}{2}=\frac{1}{4}$
$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$
Câu 1 :
\(\frac{1}{5}:\frac{3}{6}=\frac{1}{5}\times\frac{6}{3}\)
\(=\frac{6}{15}=\frac{2}{5}\)
Câu 2 :
\(\frac{4}{8}+\frac{2}{4}=\frac{1}{2}+\frac{1}{2}\)
\(=1\)
Câu 3 :
\(\frac{2}{5}\times\frac{6}{4}=\frac{12}{20}=\frac{6}{10}\)
Câu 4 :
\(\frac{7}{6}-\frac{2}{3}=\frac{7}{6}-\frac{4}{6}\)
\(=\frac{3}{6}=\frac{1}{2}\)
\(\frac{12}{1}:\frac{6}{5}+\frac{4}{7}=\frac{60}{6}+\frac{7}{4}=\frac{10}{1}+\frac{7}{4}=\frac{40}{4}+\frac{7}{4}=\frac{47}{4}\)
\(\left(\frac{5}{6}+\frac{3}{4}\right)\cdot\frac{8}{19}=\frac{38}{24}\cdot\frac{18}{9}=\frac{38}{24}\cdot2=\frac{76}{24}=\frac{19}{6}\)
\(\left(\frac{7}{4}-\frac{5}{3}\right)\cdot\frac{2}{3}=\frac{1}{12}\cdot\frac{2}{3}=\frac{1}{13}\)
15: A= 1/3-3/4+3/5+1/2007-1/36+1/15-2/9
Sửa đề:
A=-3/4-2/9-1/36+1/3+3/5+1/15+1/2007
=-27/36-8/36-1/36+5/15+9/15+1/15+1/2007
=-1+1+1/2007=1/2007
16:
\(A=\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)
=1/64
17:
=1/2-1/2+2/3-2/3+3/4-3/4+4/5-4/5+5/6-5/6-6/7
=-6/7
Cho P = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27
P x 2 = 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 ( Lấy dưới trừ trên )
P x 2 - P = 28 - 1
P x ( 2 - 1) = 28 -1
P x 1 = 28 -1
P = 28 - 1 : 1
P = 28 - 1
Tính tổng hay CM