17 - 2x = -15

2x = 17 - (-15)

2x = 32

x = 32 : 2

x = 16

--------

54 : (x + 2) = -6

x + 2 = 54 : (-6)

x + 2 = -9

x = -9 - 2

x = -11

--------

12.(3 - x) = 72

3 - x = 72 : 12

3 - x = 6

x = 3 - 6

x = -3

-------

-3(x + 5) + 18 = -27

-3(x + 5) = -27 - 18

-3(x + 5) = -45

x + 5 = -45 : (-3)

x + 5 = 15

x = 15 - 5

x = 10

-------

(x + 5)² = 9

x + 5 = 3 hoặc x + 5 = -3

*) x + 5 = 3

x = 3 - 5

x = -2

*) x + 5 = -3

x = -3 - 5

x = -8

Vậy x = -8; x = -2

--------

(3 - x)³ = 27

(3 - x)³ = 3³

3 - x = 3

x = 3 - 3

x = 0

=00

- 3\(x\) + 24  = -24

  3\(x\)          = 24 + 24

   3\(x\)          = 48

     \(x\)           = 48 : 3

     \(x\)           = 16

\(\dfrac{6}{x}=\dfrac{24}{x-7}\left(x\ne0;x\ne27\right)\)

suy ra

`6(x-27)=24x`

`=>6x-162-24x=0`

`=>-18x=162`

`=>x=-9(tm)`

\(\dfrac{6}{x}=\dfrac{24}{x-27}\left(dkxd:x\ne0,x\ne27\right)\)

\(\Leftrightarrow\dfrac{6}{x}-\dfrac{24}{x-27}=0\)

\(\Leftrightarrow\dfrac{6\left(x-27\right)-24x}{x\left(x-27\right)}=0\)

\(\Leftrightarrow6x-162-24x=0\)

\(\Leftrightarrow-18x=162\)

\(\Leftrightarrow x=--9\)

`6/x = 24/(x-27)`

`=> 6 (x-27)= 24x`

`=> 6x - 162=24x`

`=> 6x - 24x=162`

`=>-18x=162`

`=>x=162 :(-18)`

`=>x=-9`

\(\frac{x}{24}+\frac{x-6}{32}=\frac{2x-6}{27}\)

\(=>\frac{4x}{96}+\frac{3\left(x-6\right)}{96}=\frac{2x-6}{27}\)

\(=>\frac{4x}{96}+\frac{3x-18}{96}=\frac{2x-6}{27}\)

\(=>\frac{4x+3x-18}{96}=\frac{2x-6}{27}\)

\(=>\frac{7x-18}{96}=\frac{2x-6}{27}\)

=>(7x-18).27=(2x-6).96

=>189x-486=192x-576

=>3x=90=>x=30

Vậy x=30

Mày nhìn cái chóa j

12 x 50 + 8 x 27 + 6 x 52 + 24 x 53

= 24 x 25 + 24 x 9 + 24 x 13 + 24 x 53

= 24 x ( 25 + 9 + 13 + 53 )

=24 x100

= 2400

12 x 50 + 8 x 27 + 6 x 52 + 24 x 53 

=24 x 25 + 24 x 9 +24 x 13 + 24 x 53

=24 x(25 + 9 + 13 + 53)

=24x100

=2400

\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}=\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{24}\left(x^3+1\right)+x^{18}\left(x^3+1\right)+x^{12}\left(x^3+1\right)+x^6\left(x^3+1\right)+\left(x^3+1\right)}\)

=\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\frac{1}{x^3+1}\)

1120128

Tìm x εIN biết
a) 390 - (x-8) = 168:13
b) (x-140) : 7 = 27 - 24
c) x- 6 :2 - ( 48 - 24 ) :2 :6 - 3 = 0
d) x+5.2-(32+16.3:6-15)=0

b) \(\left(x-140\right):7=27-24\)

\(\left(x-140\right):7=3\)

\(x-140=21\)

\(x=161\)

      vay   \(x=161\)

c) \(x-6:2-\left(48-24\right):2:6-3=0\)

\(x-3-24:2:6-3=0\)

\(x-3-2-3=0\)

\(x-8=0\)

\(x=8\)

vay \(x=8\)

d) \(x+5.2-\left(32+16.3:6-15\right)=0\)

\(x+10-\left(32+8-15\right)=0\)

\(x+10-25=0\)

\(x-15=0\)

\(x=15\)

vay \(x=15\)

a) \(390-\left(x-8\right)=168:13\)

\(390-x+8=\frac{168}{13}\)

\(x+8=390-\frac{168}{13}\)

\(x+8=\frac{5070}{13}-\frac{168}{13}\)

\(x+8=\frac{4902}{13}\)

\(x=\frac{4902}{13}-8\)

\(x=\frac{4798}{13}\)

vay \(x=\frac{4798}{13}\)

(6^2+7^2+8^2+9^2+10^2)-(1^2+2^2+3^2+4^2+5^2)

Xét \(x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1\)

\(=\left(x^{27}+x^{21}+x^{15}+x^9+x^3\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)

\(=x^3\left(x^{24}+x^{18}+x^{12}+x^6+1\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)

\(=\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)

Vậy ta có

\(VT=\dfrac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\dfrac{1}{x^3+1}\) (đpcm)

a: \(=27\left(77+15\right)+27\cdot8\)

\(=27\cdot92+27\cdot8=27\cdot100=2700\)

b: \(=75\cdot89+75\cdot9+2\cdot75=75\left(89+9+2\right)=75\cdot100=7500\)

c: \(=0.25\left(42.9-11.7+0.8\right)=0.25\cdot32=8\)

d: 738 phút=12,3 giờ