1+4+7+10+...+67( 23 số hạng )

= ( 1 + 67 ) x 23 : 2

=782

pạn ghi rõ đề pài giúp mk tìm gì thì mk mới giúp đc pạn chứ

Áp dụng công thuwsctinhs dãy số ta có:

\(1+4+7+...+67=\frac{\left[\left(67-1\right):3+1\right].\left(67+1\right)}{2}=\frac{23.68}{2}=782\)

Ý bạn là mỗi phép tính phải thêm dấu và ....để kết quả bằng 6 phải không ạ

(1+1+1)×2=6

2+2+2=6

3×3⁰+3=6

(4+4+4)÷2=6

(5⁰+5⁰+5⁰)×2=6

6×6⁰×6⁰=6

7-7⁰×7⁰=6

8-(8⁰+8⁰)=6

(9⁰+9⁰+9⁰)×2=6

(10⁰+10⁰+10⁰)×2=6

Bài 1:

$10+3(x-6)=5^{10}:5^8=5^2=25$

$3(x-6)=25-10=15$

$x-6=15:3=5$

$x=5+6=11$

Bài 2:

a. $100-[150-8(7-4)^2]=100-(150-8.3^2)=100-150+8.3^2$

$=-50+72=72-50=22$
b. $=-999-23+999-10-67=(-999+999)-10-(67+23)$

$=0-10-90=-(10+90)=-100$

a, A = (\(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\)) - (\(\dfrac{79}{67}-\dfrac{28}{41}\))

= \(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)

= \(\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
= \(\dfrac{1}{3}-1+1\)

= \(\dfrac{1}{3}\)

@Mai Tran

`4/7 + 2/3 = 12/21 + 14/21 = 26/21`

`5/6 -3/4 = 10/12 - 9/12 = 1/12`

`3/10 xx 15/6 = 3/10 xx 5/2 = 3/4`

`7/6 : 5/4 = 7/6 xx 4/5 = 14/15`

`4/7 + 2/3 `

`=12/21 + 14/21`

`=28/21`

`5/6 - 3/4 `

`=20/24 - 18/24`

`=2/24`

`3/10 x 15/6`

`=3/4`

`7/6 : 5/4`

`=7/6 xx 4/5`

`=14/15`

`4/7 + 2/3= 12/21 + 14/21= 26/21`

`5/6-3/4=20/24 - 18/24=2/24=1/12`

`3/10 xx 15/6= (3xx15)/(10xx6)=45/60=3/4`

`7/6 : 5/4= 7/6 xx 4/5= 28/30=14/15`

a. -3/16

b. -79/20

tìm x 

x=-9

1) a) \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}\)

\(=\left(-0,75+\frac{1}{2}\right)\cdot\frac{3}{4}\)

\(=\left(-0,75+0,5\right)\cdot0,75\)

\(=-0,25\cdot0,75\)

\(=-0,1875\)

Vậy: \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}=-0,1875\)

b) \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}\)

\(=\frac{49}{20}-\frac{32}{5}\)

\(=\frac{49}{20}-\frac{128}{20}\)

\(=-\frac{79}{20}=-3\frac{19}{20}\)

Vậy: \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}=-3\frac{19}{20}\)

2) \(\frac{10}{3}\cdot x+\frac{67}{4}=-13,25\)

<=> \(\frac{10}{3}\cdot x+\frac{67}{4}=-\frac{53}{4}\)

<=> \(\frac{10}{3}\cdot x=-\left(\frac{53}{4}+\frac{67}{4}\right)\)

<=> \(\frac{10}{3}\cdot x=-30\)

<=> \(x=-30:\frac{10}{3}\)

<=> \(x=-30\cdot\frac{3}{10}\)

<=> \(x=-9\)

Vậy: \(x=-9\)