`N=|x-8|-5`

Vì \(|x-8| \ge 0 \forall x\)

\(<=>|x-8|-5 \ge -5 \forall x\)

  Hay \(N \ge -5 \forall x\)

Dấu "`=`" xảy ra `<=>|x-8|=0<=>x=8`

Trả lời:

4x=5y=10z

x.y.z =-320 => 4x. 5y.z =-6400

10z.10z.z=-6400

z³ =-64

=> z=-4

=> y=-40/5=-8

=> x=-40/4= -10

Trả lời:

(1-2x)3=-8 

=> 1-2x = -8/3

=> -2x= -8/3-1

=> 2x = 8/3+1

=> x= 11/6

`(1-2x).3=-8`

\(=>1-2x=\dfrac{-8}{3}\)

\(=>2x=1-\dfrac{-8}{3}=\dfrac{11}{3}\)

\(=>x=\dfrac{11}{3}:2\)

\(=>x=\dfrac{11}{6}\)

Tổng số tuổi của bố và con phải là số chia hết cho (10+3) em nhé!

Ta có : `30/100 = 3/10`

Tổng số phần bằng nhau :

`10+3=13(phần)`

Tuỏi của bố là :

`56 : 13 xx 10= 560/13(tuổi)`

Tuổi của con là :

`56-560/13 = 168/13(tuổi)`

`=>` Đề sai

h2.

Có: \(\widehat{B}=90^o\Rightarrow\widehat{A}=180^o-90^o=90^o\) ( trong cùng phía )

\(\widehat{ADC}=120^o\Rightarrow\widehat{BCD}=180^o-120^o=60^o\) 

h3.

\(\widehat{A}=\widehat{B}=90^o\)

\(\widehat{DCB}=70^o\) ( đối đỉnh )

\(\Rightarrow\widehat{ADC}=180^o-70^o=110^o\) ( trong cùng phía )

\(\widehat{CAa}=150^o\)

22/25.(9/11-3/2)+4/1

= 22/25 . ( -15/22) + 4/1

= -3.5 + 4/1

= 17/5

`22/25 . (9/11 -3/2) + 4/1`

`=22/25 . (9/11 -3/2) + 4`

`=22/25 . (18/22 - 33/22) + 4`

`=22/25 . (-15)/22 + 4`

`=(22xx(-15))/(25xx22) +4`

`= (-15)/25 + 4`

`=(-3)/5 + 4`

`=(-3)/5 + 20/5`

`= (-3+20)/5`

`=17/5`

`#LeMichael`

\(B=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....0.....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =0\)

\(B=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{125}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....0.....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)=0\)