làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2017}{8}\)

\(\Leftrightarrow\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)

\(\Rightarrow x-2017=0\Rightarrow x=2017\)

Vậy x=2017

giúp mk làm cầu b đi

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

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 \(\dfrac{5}{9}\)

= \(\dfrac{1}{3}\)

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

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

Ta có : A =\(\frac{2017}{2018}\)x \(\frac{7}{8}\)+ \(\frac{2017}{2018}\)x \(\frac{3}{8}\)- \(\frac{2017}{2018}\)x \(\frac{1}{4}\)

                = \(\frac{2017}{2018}\) x ( \(\frac{7}{8}+\frac{3}{8}-\frac{1}{4}\))

                  = \(\frac{2017}{2018}\)x 1

                    =\(\frac{2017}{2018}\)

Vậy A= : \(\frac{2017}{2018}\)

                                                                   Bài giải

\(A=\frac{2017}{2018}\text{ x }\frac{7}{8}+\frac{2017}{2018}\text{ x }\frac{3}{8}-\frac{2017}{2018}\text{ x }\frac{1}{4}\)

\(A=\frac{2017}{2018}\text{ x }\frac{1}{4}\left(\frac{7}{2}+\frac{3}{2}-1\right)=\frac{2017}{2018}\text{ x }\frac{1}{4}\text{ x }4==\frac{2017}{2018}\text{ x }1=\frac{2017}{2018}\)

\(\frac{x-9-4}{2017}+\frac{x-4-2017}{9}+\frac{x-2017-9}{4}=3\)

\(\Leftrightarrow\frac{x-13}{2017}+\frac{x-2021}{9}+\frac{x-2026}{4}=3\)

\(\Leftrightarrow\frac{x}{2017}+\frac{x}{9}+\frac{x}{4}=3+\frac{2021}{9}+\frac{2026}{4}+\frac{13}{2017}\)

\(\Leftrightarrow x\left(\frac{1}{2017}+\frac{1}{9}+\frac{1}{4}\right)=....\left(\text{tự giải tiếp :3}\right)\)

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

a/   3^21 > 2^31

b/   2017^10 +  2017^9  <2018^10

chọn mình nha . Mình cũng học lớp 6 đó (>-<)

search gg đi

Đặt:  \(x-1=a;\)\(y-3=b;\)\(z-8=c\)

=>  \(a+b+c=x+y+z-12=0\)(do  x+y+z = 12 )

Ta dễ dàng chứng minh được:

nếu  a + b + c = 0 

thì:  a³ + b³ + c³ = 3abc

Như vậy ta có:

\(\left(x-1\right)^3+\left(y-3\right)^3+\left(z-8\right)^3=0\)

<=>  \(3\left(x-1\right)\left(y-3\right)\left(z-8\right)=0\)

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