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1.
\(\left(572\cdot7+266\right)\cdot\left(366\cdot9-168\cdot18\right)\cdot\left(346\cdot6-348\right)\)
\(=\left(286\cdot7\cdot2+133\cdot2\right)\cdot\left(366\cdot9-168\cdot2\cdot9\right)\cdot\left(173\cdot6\cdot2-174\cdot2\right)\)
\(=\left(2002\cdot2+133\cdot2\right)\cdot\left(366\cdot9-336\cdot9\right)\cdot\left(1038\cdot2-174\cdot2\right)\)
\(=\left[2\cdot\left(2002+133\right)\right]\cdot\left[9\cdot\left(366-336\right)\right]\cdot\left[2\cdot\left(1038-174\right)\right]\)
\(=2\cdot2135\cdot9\cdot30\cdot2\cdot864\)
\(=4270\cdot9\cdot30\cdot2\cdot864\)
\(=\left(4270\cdot30\right)\cdot9\cdot2\cdot864\)
\(=\left(427\cdot10\right)\cdot\left(3\cdot10\right)\cdot9\cdot2\cdot864\)
\(=\left(427\cdot3\right)\cdot\left(10\cdot10\right)\cdot9\cdot2\cdot864\)
\(=1281\cdot100\cdot9\cdot2\cdot864\)
\(=\left(1281\cdot100\right)\cdot\left(9\cdot2\cdot864\right)\)
\(=\left(1281\cdot100\right)\cdot15552\)
\(=\left(1281\cdot15552\right)\cdot100\)
\(=19922112\cdot100\)
\(=1992211200\)
2.
\(\left(1+3+5+7+...+97+9\right)\cdot\left(45\cdot3-15\cdot2-45\right)\)
\(=\left(1+3+5+7+...+97+9\right)\cdot\left(15\cdot3\cdot3-15\cdot2-15\cdot3\right)\)
\(=\left(1+3+5+7+...+97+9\right)\cdot\left[15\cdot\left(3\cdot3\right)-15\cdot2-15\cdot3\right]\)
\(=\left(1+3+5+7+...+97+9\right)\cdot\left(15\cdot9-15\cdot2-15\cdot3\right)\)
\(=\left(1+3+5+7+...+97+9\right)\cdot\left[15\cdot\left(9-2-3\right)\right]\)
\(=\left(1+3+5+7+...+97+9\right)\cdot15\cdot4\)
\(=\left[\left(1+3+5+7+...+97\right)+9\right]\cdot\left(15\cdot4\right)\)
Trong \(\left(1+3+5+7+...+97\right)\) có số số hạng là:
\(\left(97-1\right)\div2+1=49\) ( số hạng )
\(\Rightarrow\left[\left(1+3+5+7+...+97\right)+9\right]\cdot\left(15\cdot4\right)\)
\(=\left[\left(97+1\right)\cdot49\div2\right]\cdot\left(15\cdot4\right)\)
\(=2401\cdot60\)
\(=\left(2401\cdot6\right)\cdot10\)
\(=14406\cdot10\)
\(=144060\)
3.
\(\left(180\div15-132\div11\right)\cdot\left(57869-297\div11\cdot108\right)\)
\(=\left(12-12\right)\cdot\left(57869-297\div11\cdot108\right)\)
\(=0\cdot\left(57869-297\div11\cdot108\right)\)
\(=0\)
Sửa lại bài 2; dòng 11 ( từ đề bài bài 2 ):
\(=\left\{\left[\left(97+1\right)\cdot49\div2\right]+9\right\}\cdot\left(15\cdot4\right)\)
\(=\left(2401+9\right)\cdot\left(15\cdot4\right)\)
\(=2410\cdot\left(15\cdot4\right)\)
\(=2410\cdot60\)
\(=\left(241\cdot10\right)\cdot\left(6\cdot10\right)\)
\(=\left(241\cdot6\right)\cdot\left(10\cdot10\right)\)
\(=1446\cdot10\)
\(=14460\)
x x - 4/7 = 15/20
(hình như sai đề, tại vì lớp 4 chưa học lũy thừa)
x * 2/7 -2/3 =1/2
x* 2/7 = 1/2 + 2/3
x* 2/7 = 7/6
x= 7/6 : 2/7
x= 49/12
x : 5/4 + 4/5 =2
x: 5/4 = 2 - 4/5
x: 5/4 = 6/5
x= 6/5 * 5/4
x= 3/2
4/3 x 7/8 x 5
= 7/6 x 5
= 35/6
8/9 x 6/5 x 2/3
= 16/15 x 2/3
= 32/45
4/7 : 5/7 x 10/11
= 4/5 x 10/11
= 8/11
8/13 : 7/13 : 4
= 8/7 : 4
= 2/7
a) 3/2 x 4/7 = 6/7
8/21 : 2/3= 4/7
8/21 : 4/7 = 2/3
4/7 x 2/3 = 8/21
b) 3/11 x 2 = 6/11
6/11: 3/11 = 2
1+2+3+4+5+6
= (1+6) x 3
= 7 x 3
= 21
11+22+33+44+55+66
= (11+66) x 3
= 77 x 3
= 231