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D
1 tháng 5 2018
\(a,\)Ta có:
\(1-\frac{167}{168}=\frac{1}{168}\)
\(1-\frac{176}{177}=\frac{1}{177}\)
Vì \(\frac{1}{168}>\frac{1}{177}\)
\(\Rightarrow\frac{167}{168}< \frac{176}{177}\)
XO
26 tháng 10 2020
Gọi số đã bớt đi là a
Ta có \(\frac{168-a}{180-a}=\frac{3}{5}\)
=> \(168-a=\frac{3}{5}\times\left(180-a\right)\)
=> \(168-a=108-\frac{3}{5}\times a\)
=> \(a-\frac{3}{5}\times a=60\)
=> \(a\times\frac{2}{5}=60\)
=> a = 150
Vậy số đã bớt đi là 150
26 tháng 10 2020
Bài giải: (mình ghi cách làm nha)
\(180-168=12\)
\(5-3=2\)(phần)
\(12:2\times3=18\)
\(168-18=150\)
16 tháng 6 2020
\(a.\frac{4}{3}-\frac{3}{2}:X=\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{4}{3}-\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{8}{6}-\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{7}{6}\)
\(X=\frac{3}{2}:\frac{7}{6}\)
\(X=\frac{3}{2}\times\frac{6}{7}\)
\(X=\frac{9}{7}\)
16 tháng 6 2020
\(b.\left(X+\frac{2}{3}\right):\frac{1}{3}=\frac{41}{3}\)
\(X-\frac{2}{3}=\frac{41}{3}.\frac{1}{3}\)
\(X-\frac{2}{3}=\frac{41}{9}\)
\(X=\frac{41}{9}+\frac{2}{3}\)
\(X=\frac{41}{9}+\frac{6}{9}\)
\(X=\frac{47}{9}\)
13 tháng 3 2020
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.
\(\left(180\div15-132\div11\right)\cdot\left(57869-297\div11\cdot108\right)\)
\(=\left(12-12\right)\cdot\left(57869-297\div11\cdot108\right)\)
\(=0\cdot\left(57869-297\div11\cdot108\right)\)
\(=0\)
13 tháng 3 2020
Sửa lại bài 2; dòng 11 ( từ đề bài bài 2 ):
\(=\left\{\left[\left(97+1\right)\cdot49\div2\right]+9\right\}\cdot\left(15\cdot4\right)\)
\(=\left(2401+9\right)\cdot\left(15\cdot4\right)\)
\(=2410\cdot\left(15\cdot4\right)\)
\(=2410\cdot60\)
\(=\left(241\cdot10\right)\cdot\left(6\cdot10\right)\)
\(=\left(241\cdot6\right)\cdot\left(10\cdot10\right)\)
\(=1446\cdot10\)
\(=14460\)
4 tháng 7 2017
bài 1:
\(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}\)
\(=\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}\)
\(=\frac{11}{11}+\frac{1}{3}=1+\frac{1}{3}=\frac{3}{3}+\frac{1}{3}=\frac{4}{3}\)
bài 2:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{12}\right)\)
\(=\frac{11}{20}+\frac{1}{4}=\frac{11}{20}+\frac{5}{20}=\frac{15}{20}=\frac{3}{4}\)
bài 3:
a) \(\frac{3}{2}\cdot\frac{4}{5}\cdot\frac{2}{3}=\left(\frac{3}{2}\cdot\frac{2}{3}\right)\cdot\frac{4}{5}=1\cdot\frac{4}{5}=\frac{4}{5}\)
b) \(\frac{6}{7}\cdot\frac{5}{3}\cdot\frac{7}{6}=\left(\frac{6}{7}\cdot\frac{7}{6}\right)\cdot\frac{5}{3}=1\cdot\frac{5}{3}=\frac{5}{3}\)
bài 4:
a) \(\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{4}\cdot\frac{2}{5}=\frac{2}{5}\cdot\left(\frac{1}{4}+\frac{3}{4}\right)=\frac{2}{5}\cdot1=\frac{2}{5}\)
b) \(\frac{6}{11}:\frac{2}{3}+\frac{5}{11}:\frac{2}{3}=\left(\frac{6}{11}+\frac{5}{11}\right):\frac{2}{3}=1:\frac{2}{3}=\frac{3}{2}\)
4 tháng 7 2017
Bài 1:
6/11 + 1/3 + 5/11
= ( 6/11 + 5/11) + 1/3
= 11/11 + 1/3
= 1 + 1/3
= 3/3 +1/3
= 4/3
Bài 2:
1/2 + 1/6 + 1/12 + 1/20
= ( 1/2 + 1/6 + 1/12 ) + 1/20
= ( 6/12 + 2/12 + 1/12 ) + 1/20
=9/12 + 1/20
= 3/4 +1/20
= 15/20 + 1/20
= 16/20 = 4/5
Bài 3:
a) \(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{3}\) \(=\left(\frac{3}{2}\times\frac{2}{3}\right)\times\frac{4}{5}\)\(=1\times\frac{4}{5}=\frac{4}{5}\)
b) \(\frac{6}{7}\times\left(\frac{5}{3}\times\frac{7}{6}\right)\) \(=\frac{6}{7}\times\frac{35}{18}\)\(=\frac{1\times5}{7\times3}=\frac{5}{21}\)
Bài 4:
a) 2/5 x 1/4 + 3/4 x 2/5
= 2/5 x ( 1/4 + 3/4)
= 2/5 x 1
= 2/5
b) 6/11 : 2/3 +5/11 : 2/3
= ( 6/11 + 5/11) x 3/2
= 11/11 x 3/2
= 1 x 3/2
= 3/2
....
6 tháng 7 2017
\(a.\left(\frac{6}{11}+\frac{5}{11}\right).\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}b.\frac{3}{5}\cdot\frac{7}{9}+\frac{3}{5}\cdot\frac{2}{9}=\frac{3}{5}\cdot\left(\frac{7}{9}+\frac{2}{9}\right)=\frac{3}{5}\cdot1=\frac{3}{5}\)
504/11
504/11