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a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=\frac{15}{15}+\frac{7}{7}\)
\(=1+1\)
\(=2\)
b) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)
a, = 15 x (3+7) + 15 = 15 x 10 + 15 = 150 + 15 = 165
b, = 24 x 18 - 24 x 8 + 24 = 24 x (18-8) + 24 = 24 x 10 + 24 = 240 + 24 = 264
k mk nha
a,15*3+15*7+15
=15*(3+7)
=15*10
=150
b,24*18-24*7
=24*(18-7)
=24*11
=264
tk cho mk nha^-^
(18/1+12/15)+7/15=2+7/15=37/15
19/37+1/37-19/37=1/37
26/7\(\times\)(11/3-8/3)=26/7\(\times\)1=26/7
1x1x1x1/3x3x2=1/18
cho mình xin 2 k
b)1/2*2/3*3/4*4/5*5/6*6/7*7/8*8/9*9/10
=1x2x3x4x5x6x7x8x9
2x3x4x5x6x7x8x9x10
= 1
10
a) \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)
\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)
\(=\frac{1}{2}+\frac{1}{3}\)
\(=\frac{5}{6}\)
b) \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)
\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)
\(=1\times\frac{1}{2}=\frac{1}{2}\)
c) \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)
\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)
\(=\frac{247}{420}+\frac{72}{420}\)
\(=\frac{319}{420}\)
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 ):
\(=\left\{\left[\left(97+1\right)\cdot49\div2\right]+9\right\}\cdot\left(15\cdot4\right)\)
\(=\left(2401+9\right)\cdot\left(15\cdot4\right)\)
\(=2410\cdot\left(15\cdot4\right)\)
\(=2410\cdot60\)
\(=\left(241\cdot10\right)\cdot\left(6\cdot10\right)\)
\(=\left(241\cdot6\right)\cdot\left(10\cdot10\right)\)
\(=1446\cdot10\)
\(=14460\)
15/17.(4/7+3/7)-1/3=15/17.1-1/3=15/17-1/3=28/51