\(\frac{41^2-39}{41^2+39^2+82.39}\) TÍNH NHANH
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\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
a) \(413.\left(413-26\right)+169=413^2-2.13.413+13^2=\left(413-13\right)^2=160000\)
b) \(\left(625^2+3\right).\left(25^4-3\right)-5^{16}+10\)
\(=\left(5^8+3\right)\left(5^8-3\right)-5^{16}+10\)
\(=5^{16}-9-5^{16}+10=1\)
c) \(\frac{41^2+39^2+8^2.39}{41^2-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)
a) 41. 36+ 59. 90 + 41. 84 + 59. 30
= 41 . (36 + 84) + 59 . (90 + 30)
= 41 . 120 + 59 . 120
= 120 . ( 41 + 59 )
= 120 . 100
= 12000
b) 4. 51 . 7 + 2 . 86 .7 + 6 . 4 . 7
= 28 . 51 + 2 . 43 . 2 . 7 + 6 . 4 . 7
= 28 . 51 + 28 . 43 + 28 . 6
= 28 . ( 51 + 43 + 6 )
= 28 . 100
= 2800
c) ( 4.7 ) . 5 + ( 4.7 ) . 43 + ( 4.7) - 6
= 28 . 5 + 28 .43 + 28 - 6
= 28 . (5 + 43 + 1) - 6
= 28 . 49 - 6
= 1372 - 6
= 1366.
d) 53 . 39 + 47 . 39 - 53 . 21 - 47 . 21
= 53 . (39 - 21) + 47. (39 - 21)
= (39 - 21) . ( 53 + 47 )
= 18 . 100
= 1800.
Theo đề bài :
Vì số cuối là 3072 nên tổng các số trên là 3072
( 3072 + 1 ) x 3072 : 2 = 4720128
Đáp số: 4720128
~~Hok tốt~~
a. -3752 - (39 - 3632) - 41
= -3752 - -3593 - 41
= -159 - 41
= -200
b. - (-2023 - 80 + 94) - (2023 + 80 - 91)
= -2037 - 2012
= -4049
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a/ = 39 x ( 39 + 41 + 20 )
= 39 x 100 = 3900
b/ = 428 x ( 75 + 5 - 70 )
= 428 x 10 = 4280
a) 39 x 39 + 41 x 39 + 39 x 20
= 39 x (39 + 41 + 20)
= 39 x 100
= 3900
b) 428 x 75 + 5 x 428 - 70 x 428
= 428 x (75 + 5 - 70)
= 428 x 10
= 4280
39 x 57 + 39 x 41 + 39 x 2
= 39 x (57 + 41 + 2)
= 39 x 100
= 3900
\(=\frac{41^2-39}{41^2+39^2+2.41.39}\)
\(=\frac{41^2-39}{41^2+2.41.39+39^2}\)
\(=\frac{41^2-39}{\left(41+39\right)^2}\)
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