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mk chi chep thua so thu 3 thoi nhe con 2 thua so con lai giu nguyen
42/30.75/23-19/23.210/38
=7/5.75.1/23-1/23.19.210/38
=7.15.1/23-1/23.210/2
=105.1/23-1/23.105=0
gia tri bieu thuc=0
minh cung dang thac mac cau nay ban nao giup minh voi
a)
13 x 81 + 13 x 19
= 13 x ( 81 + 19 )
= 13 x 100 = 1300
b)
246 x 83 + 246 x 17
= 246 x ( 83 + 17 )
= 246 x 100
= 24600
c)
246 x 83 + 247 x 17
= 246 x 83 + 246 x 17 + 17
= 246 x ( 83 + 17 ) + 17
= 246 x 100 + 17
= 24600 + 17
= 24617
Ủng hộ mik nhak !!!
13x81+13x19
=13x(19+81)
=13x100
=1300
246x83+246x17
=246x(83+17)
=246x100
=24600
a, 42x(54+17):71
=42x71:71=42
b, 123x(154-65):89
=123x89:89=123
c,324x(6+4):324
=324x10:324=10x324:324=10
\(a)\left(42\times54+17\times42\right):71\)
\(=42\times54+17\times42:71\)
\(=42\times\left(54+17\right):71\)
\(=42\times71:71\)
\(=42\)
\(b)\left(123\times154-65\times123\right):89\)
\(=123\times154-65\times123:89\)
\(=123\times\left(154-65\right):89\)
\(=123\times89:89\)
\(=123\)
\(c)\left(324\times6+4\times324\right):\left(162\times2\right)\)
\(=324\times6+4\times324:324\)
\(=324\times\left(6+4\right):324\)
\(=324\times10:324\)
\(=3240:324\)
\(=10\)
Bài 1:
a) x - 32 + 25 = 12 x 2
x- 32 +25 = 24
x -32 = -1
x = 31
b) 24 + x - 3 = 21
24+x = 24
x = 0
c) x * 10 : 2 = 40
x *10 = 80
x = 8
Bài 2:
a) 123 - 82 x 6= 123- 492= -369
b) 458 - 125 + 23= 333+23 = 356
c) 325 + 125 x 125 x 325= 325 + 5 078 450
Cái này giống như bài lớp 6 ý.
Bài 1 : Tìm x
a) x - 32 + 25 = 12 x 2
x - 32 + 25 = 24
x - 32 = 24 - 25
x - 32 = ( - 1 )
x = ( -1 ) + 32
x = 31
b) 24 + x - 3 = 21
24 + x = 21 + 3
24 + x = 24
x = 24 - 24
x = 0
c) x * 10 : 2 = 40
x * 10 = 40 x 2
x * 10 = 80
x = 80 : 10
x = 8
Bài 2: Tính giá trị của biểu thức
a) 123 - 82 x 6
123 - 492
-369
b) 458 - 125 + 23
333 + 23
356
c) 325 + 125 x 125 x 325
325 + 5078125
5078450
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
Ta thấy tích 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 có chứa thừa số 2 và 5 nên tích này có tận cùng là 0, không thể = 3999
=> Huệ tính sai
ta có 2x3x5x7x11x13x17x19x23x31x37
có hai thưa số 2 va 5 .cho nên hs nay có tận cung la 0
vậy không thể = 3999
vậy phép tính của huệ sai nha
123 x 17 + 123 x 23 + 123 x 2 x 30
123 x (17 + 23 + 2 x 30)
123 x 100
12300
123 x 17 + 123 x 23 + 246 x 30
= 123 x (17 + 23) + 246 x 30
= 123 x 40 + 246 x 30
= 123 x 2 x 20 + 246 x 30
= 246 x 20 + 246 x 30
= 246 x (20 + 30)
= 50 x 246
= 12300
~Học tốt~