\(a,34\times3+28\\ =102+28\\ =130\\ b,223-23\times5\\ =223-115\\ =108\\ c,354:6-264:12\\ =59-22\\ =37\\ d,\left(18+37\right)\times24\\ =55\times24\\ =1320\\ e,\left(256+64\right):16\\ =320:16\\ =20\\ g,\left(121-88\right):11\\ =33:11\\ =3\)

\(a,150+?=213\\ ?=213-150\\ ?=63\)

\(b,56\times?=560\\ ?=560:56\\ ?=10\\ 360-?=259\\ ?=360-259\\ ?=101\\ 8700:?=87\\ ?=8700:87\\ ?=100\\ ?-81=265\\ ?=265+81\\ ?=346\\ ?:1000=79\\ ?=79\times1000\\ ?=79000\)

\(105-\left\{5\cdot5-\left[\left(2^4+8^2\right):4\right]\right\}\\ =105-\left\{25-\left[80:4\right]\right\}\\ =105-\left\{25-20\right\}\\ =105-5\\ =100\\ \left\{5+\left[2^4+8^2:\left(9-5\right)\right]\right\}\cdot2\\ =\left\{5+\left[16+64:4\right]\right\}\cdot2\\ =\left\{5+\left[16+16\right]\right\}\cdot2\\ =37\cdot2\\ =74\\ 132-\left\{51-\left[2^3\cdot\left(8-3\right)+1\right]\right\}\\ =132-\left\{51-\left[8\cdot5+1\right]\right\}\\ =132-\left\{51-41\right\}\\ =132-10\\ =122\)

Nếu 45 xe ô tô như thế thì chở được là:

\(37,6:9\times45=188\) (tấn) \(=188000kg\)

Đáp số: \(188000kg\)

\(2\cdot4\cdot8\cdot8\cdot8\cdot8\cdot2\cdot2\cdot2=2\cdot2^2\cdot2^3\cdot2^3\cdot2^3\cdot2^3\cdot2\cdot2\cdot2=2^{1+2+3+3+3+3+1+1+1}=2^{18}\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{15\times16}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{16}\\ =1-\dfrac{1}{16}\\ =\dfrac{16}{16}-\dfrac{1}{16}\\ =\dfrac{15}{16}\)

\(\dfrac{7}{12}:\dfrac{7}{9}=\dfrac{7}{12}\times\dfrac{9}{7}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{2007\times2009}+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{2007\times2009}+\dfrac{2}{2009\times2011}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{2011}\right)\\ =\dfrac{1}{2}\times\dfrac{2010}{2011}\\ =\dfrac{1005}{2011}\)

Tỉ số đường chéo thứ nhất và đường chéo thứ hai là: \(\dfrac{2}{1}\)

Tổng số phần bằng nhau là:

\(2+1=3\) (phần)

Đường chéo thứ nhất là:

\(411:3\cdot2=274\left(m\right)\)

Đường chéo thứ hai là:

\(411-274=137\left(m\right)\)

Diện tích hình thoi là:

\(\dfrac{274\cdot137}{2}=18769\left(m^2\right)\)

Đáp số: \(18769m^2\)

\(9673=9\cdot10^3+6\cdot10^2+7\cdot10^1+3\cdot10^0\\ 3547=3\cdot10^3+5\cdot10^2+4\cdot10^1+7\cdot10^0\\ abcde=a\cdot10^4+b\cdot10^3+c\cdot10^2+d\cdot10^1+e\cdot10^0\)