Từ trang `1` đến trang `9` có `9` số `->9` chữ số

Từ trang `10` đến trang `99` có `90` số `->180` chữ số

Còn lại số chữ số : `402-9-180=213` (chữ số)

Cuốn sách có : `99+213:3=170` (trang)

(x-1)⁴=16

(x-1)⁴=2⁴

⇒x-1=2

      x=2+1

      x=3

x = 3

8 . 9 . 14 + 6 . 17 . 12 + 19 . 4 . 18

= 72 . 14 + 72 . 17 + 72 . 19

= 72 . ( 14 + 17 + 19 )

= 72 . 50

= 3600  

$8\times 9\times 14+6\times 17\times 12+19\times 4\times 18$

$=72\times 14+72\times 17+72\times 19$

$=72\times (14+17+19)$

$=72\times 50$

$=3600$

\(=\dfrac{4}{25}+\dfrac{1}{2}\left(\dfrac{9}{2}-2\right)-\dfrac{1}{4}=\dfrac{4}{25}+\dfrac{1}{2}\left(\dfrac{5}{2}\right)-\dfrac{1}{4}=\dfrac{4}{25}+\dfrac{5}{4}-\dfrac{1}{4}=\dfrac{4}{25}+1=\dfrac{29}{25}\)

`(-2/5)^2 + 1/2 . (4,5 - 2) - 25%`

`=4/25 + 1/2 . (9/2 - 2) - 1/4`

`=4/25 + 1/2 . (9/2 - 4/2)-1/4`

`=4/25 + 1/2 . 5/2 - 1/4`

`=4/25 + 5/4 - 1/4`

`=4/25 + 1`

`=4/25  + 25/25`

`=29/25`

a] 425 . 7 . 4 - 170 . 60

= ( 425 . 4 ) - ( 7 .170 ) . 60

= 1700 - 1190 . 60

= 510 . 60

= 30600

a] 425 . 7 .4 - 170 . 60

= ( 425 . 4 ) . 7 - ( 170 . 10 ) . 6

= 1700 . 7 - 1700 . 6

= 1700 . ( 7-6 )

= 1700 . 1

= 1700 

a] 17 . 82 + 19 . 17 - 17

= 17 . 82 + 19 . 17 - 17.1

= 17 . ( 82 + 19  - 1 )

= 17 . 100

= 1 700 

a] 17 . 82 + 19 . 17 - 17

= 17 . ( 82 + 19 - 17 )

= 17 . 84

= 1428

a, \(A=\left(2011-1\right)\left(2011+1\right)=2011^2-1< B\)

b, \(C=\left(2012-2\right)\left(2012+2\right)=2012^2-4< D\)

.

a] 28 . 62 - 28 . 57

= 28 . ( 62 - 57 )

= 28 . 5

= 140 

a] 28 . 62 - 28 . 57

= 28 . ( 62 - 57 )

= 28 . 5

= 140 

a, \(\left(x+1\right)\left(x+2\right)=1.12=2.6=3.4\)

TH1 \(\left\{{}\begin{matrix}x+1=1\\x+2=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

TH2 \(\left\{{}\begin{matrix}x+1=2\\x+2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

TH3 : \(\left\{{}\begin{matrix}x+1=3\\x+2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)

b, \(x^2+x-12=0\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)

c, \(\dfrac{x+5}{x}=1+\dfrac{5}{x}\Rightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)