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\(=\left(2x+\frac{3}{4}\right)\frac{7}{9}=\frac{15}{8}\)
\(=2x+\frac{3}{4}\)\(=\frac{15}{8}:\frac{7}{9}\)
=\(2x+\frac{3}{4}=\frac{135}{56}\)
=2x=\(\frac{135}{56}-\frac{3}{4}\)
=2x=\(\frac{93}{56}\)
x=\(\frac{93}{56}:2\)
x=\(\frac{93}{112}\)
k nha
Ta có:
abcabc = abc x 1000 + abc
= abc x 1001
= abc x 7 x 11 x 13 chia hết cho 7; 11; 13
Chứng tỏ ...
abcabc = 1000abc + abc = 1000abc + 1abc = 1001abc chia het cho 7,11,13(vi 1001 chia het cho 7,11,13)
\(10^6-5^7=\left(2.5\right)^6-5^7=2^6.5^6-5^7=5^6\left(2^6-5\right)\)
\(=5^6\left(64-5\right)=5^6.59\)chia hết cho 59(đpcm)
Gọi tập là A, Ta có:
A = { -4 ; -3 ; -2 ; -1 ; 0; 1; 2 ; 3; 4; 5}
Tổng của A là:
(-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5
=[(-4) + (-3) + (-2) + (-1)] + 0 + [1 + 2 + 3 + 4 + 5]
= (-10) + 15
=(15 - 10)
= 5
2/35 + 4/77 + 2/143 + 4/221 + 2/323 + 4/437 + 2/575
= 2/5.7 + 4/7.11 + 2/11.13 + 4/13.17 + 2/17.19 + 4/19.23 + 2/23.25
= 1/5 - 1/7 + 1/7 - 1/11 + 1/11 - 1/13 + 1/13 - 1/17 + 1/17 - 1/19 + 1/19 - 1/23 + 1/23 - 1/15
= 1/5 - 1/25
= 4/25
\(\frac{2}{35}\)+ \(\frac{4}{77}\)+ \(\frac{2}{143}\)+ \(\frac{4}{221}\)+ \(\frac{2}{323}\)+ \(\frac{4}{437}\)+ \(\frac{2}{575}\).
= \(\frac{2}{5.7}\)+ \(\frac{4}{7.11}\)+ \(\frac{2}{11.13}\)+ \(\frac{4}{13.17}\)+ \(\frac{4}{17.21}\)+ \(\frac{2}{21.23}\).
= \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{11}\)+ \(\frac{1}{11}\)- \(\frac{1}{13}\)+ \(\frac{1}{13}\)- \(\frac{1}{17}\)+ \(\frac{1}{17}\)- \(\frac{1}{21}\)+ \(\frac{1}{21}\)- \(\frac{1}{23}\).
= \(\frac{1}{5}\)- \(\frac{1}{23}\).
= \(\frac{23}{115}\)- \(\frac{5}{115}\).
= \(\frac{18}{115}\).
\(2n+3⋮3n+4\Leftrightarrow6n+9⋮3n+4\)
\(\Leftrightarrow2\left(3n+4\right)+1⋮3n+4\Leftrightarrow1⋮3n+4\)
\(\Rightarrow3n+4\inƯ\left(1\right)=\left\{\pm1\right\}\)
3n + 4 | 1 | -1 |
3n | -3 | -5 |
n | 1 | -5/3 |
\(2n+3⋮3n+4\)
Ta có: \(2n+3=3\left(2n+3\right)=6n+9\)
\(3n+4⋮3n+4\Leftrightarrow2\left(3n+4\right)⋮3n+4\Leftrightarrow6n+8⋮3n+4\Leftrightarrow\left(6n+9\right)-\left(6n+8\right)⋮3n+4\)
\(\Leftrightarrow1⋮3n+4\Leftrightarrow3n+4\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow n\in\left\{-1;\frac{-5}{3}\right\}\)
-7-4=-11
-11