\(a,2017\times\left(8+2\right)=2017\times10=20170\\ b,=\dfrac{3}{7}\times\left(\dfrac{9}{7}-\dfrac{2}{7}\right)=\dfrac{3}{7}\times1=\dfrac{3}{7}\)

a) 2017 x 34 + 2017 x 65 + 2017

= 2017 x ( 34+65+1)

= 2017 x 100

= 201700

b) 3/5 x 7/9 - 3/5 x 2/9

= 3/5 x ( 7/9- 2/9)

= 3/5 x 5/9

= 1/3

a) 2017 x 34 + 2017 x 65 + 2017

= 2017 x (34 + 65 + 1)

= 2017 x 100

= 201700

b) \(\dfrac{3}{5}\) x \(\dfrac{7}{9}\) - \(\dfrac{3}{5}\) x \(\dfrac{2}{9}\)

= \(\dfrac{3}{5}\) x \(\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\)

= \(\dfrac{3}{5}\) x \(\dfrac{5}{9}\)

= \(\dfrac{1}{3}\)

\(a,=2017\times\left(34+65+1\right)=2017\times100=201700\)

\(b,\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)=\dfrac{3}{5}\times\dfrac{5}{9}=\dfrac{15}{45}=\dfrac{1}{3}\)

32 x 27 + 148 x 34 + 16 x 94

= ( 32 x 27 + 16 x 94 ) + 74 * 2 * 34

= 32 ( 27 + 47 ) + 74 x 68

= 32 x 74 + 74 x 68

= ( 32 + 68 ) x 74

= 100 * 74

= 7400

\(\frac{2017\cdot2013-2017}{4034\cdot1006+2007}\)

\(=\frac{2017\cdot\left(2013-1\right)}{2017\cdot\left(2012+1\right)}\)

\(=\frac{2012}{2013}\)

a, 32×27+148×34+16×94

=32×27+74×2×34+16×2×47

=32×27+74×68+32×47

=32×(27+47)+74×68

=32×74+74×68

=74×(32+68)

=72×100

=7200

b, 2017×2013-2017 / 4034×1006+2017

=2017×(2013-1) / 2017×2×1006+2017

=2017×2012 / 2017×2012+2017

=1/2017

a) 2017 x 34 + 2017 x 65 + 2017

= 2017 x ( 34 + 65 + 1 )

= 2017 x 100

= 201700

b) 2007 - ( 0 x 3/4 x 1/4 )

= 2007 - 0

= 2007

Chúc bạn học tốt!

\(a)2017x34+2017x65+2017\)

\(=2017x\left(34+65+1\right)\)

\(=2017x100\)

\(=201700\)

\(b)2007-\left(0x\frac{3}{4}x\frac{1}{4}\right)\)

\(=2007-0\)

\(=2007\)

Chúc bạn học tốt !!! 

thank mấy bạn

\(\frac{0,5+\frac{2}{43}-\frac{2}{2017}}{\frac{3}{4}+\frac{3}{43}-\frac{3}{2017}}:\frac{\frac{4}{91}+\frac{4}{34}+\frac{4}{2017}}{\frac{5}{91}+\frac{5}{34}+\frac{5}{2017}}\)

\(=\frac{\frac{2}{4}+\frac{2}{43}-\frac{2}{2017}}{\frac{3}{4}+\frac{3}{43}-\frac{3}{2017}}:\frac{4\left(\frac{1}{91}+\frac{1}{34}+\frac{1}{2017}\right)}{5\left(\frac{1}{91}+\frac{1}{34}+\frac{1}{2017}\right)}\)

\(=\frac{2\left(\frac{1}{4}+\frac{1}{43}-\frac{1}{2017}\right)}{3\left(\frac{1}{4}+\frac{1}{43}-\frac{1}{2017}\right)}:\frac{4}{5}\)

\(=\frac{2}{3}.\frac{5}{4}=\frac{5}{6}\)