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 \(\dfrac{5}{9}\)

= \(\dfrac{1}{3}\)

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

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

Tính nhanh:

a) 36.(65-79)+65|35-79|

=36.(-14) + 65.44

=-504 + 2860

=2356

b) 12.(34-86)+34.(86-12)+86.(12-34)-12.(34+86)

= 12.(-52) +34.74+86. (-22) - 12.120

=-624 + 2516 +(-1892) -1440

= 1892 + (-1892) -1440

=0- 1440

=-1440

a) 36 . (-14) + 65 . 44

= -504 + 2860

= 2356

Mình làm cho bạn câu cuối

...(x-3)+(x-2)+(x-1)+0+1+2+...+11=11

VP =11

=> VT=11

Để VT =11 cần có đủ cặp số đối đến (10)=> x-3=-10=> x=-7

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

Ta có : \(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)

\(\Rightarrow\left(4x^2+y^2+z^2-4xy-4zx+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

\(\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\forall x,y,z\\\left(y-3\right)^2\ge0\forall y\\\left(z-5\right)^2\ge0\forall z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(2x-y-z\right)^2=0\\\left(y-3\right)^2=0\\\left(z-5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-3-5=0\\y=3\\z=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=8\\y=3\\z=5\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\left(1\right)\)

Lại có : \(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)

Thay \(\left(1\right)\)vào \(S\),ta được :

\(S=0^{2017}+\left(-1\right)^{2017}+1^{2017}\)

    \(=0-1+1=0\)

Vậy \(S=0\)

a) (17.x - 25 ) : 8 = 92 : 65= 92/65

17x - 25 = 92/65.8

17x = 2361/65

x= 2361/65 : 17=2361/1105

b) x- 60 = 2017¹⁰¹ - 2017¹⁰⁰

   x-60 = 2017

        x = 2017 + 60

        x = 2077

c) 5x . 30-3x.x = 45

   5x.3x.x   = 45 + 30 = 75

   15x.x      = 75

B) 2017¹⁰⁰.(x-60)=2017¹⁰¹

x-60=2017¹⁰¹ : 2017¹⁰⁰

x-60 = 2017

x= 2017 + 60

x= 2077

C) 5.x+3.(10-x)+x=45

 5x+30 - 3x + x = 45

3x = 45-30

3x = 15

x= 5

A) (17x-25):8+65=9²

(17x-25):8 = 81 - 65

(17x-25):8 = 16

17x-25= 16:8

17x = 2+27

17x = 29

x= 29/17

2017 x 7 - 2017 x 4 - 2017 x 3 + 2017

=2017.(7-4-3)

=2017.0

=0

\(2017\times7-2017\times4-2017\times3+2017\)

\(=2017\times(7-4-3+1)\)

\(=2017\times1=2017\)

2017*7 - 2017*4 - 2017*3 +2017 = 2017