2- X x 27/11 = 35/22

    X x 27/11 =2-35/22

   X x 27/11  =9/22

            X     =9/22:27/11

            X    =1/6

\(4^{20}+4^{21}+4^{22}+4^{23}=4^{20}\left(1+4+4^2+4^3\right)=4^{20}\cdot85⋮5\left(85⋮5\right)\)

\(=5\cdot\left(4^{20}+4^{22}\right)⋮5\)

0,25 × X - 27/8 × X = 3/4

X × ( 0,25 - 27/8 ) = 0,75

X × ( -25/8 ) = 0,75

X = 0,75 ÷ ( -25/8 )

X = -6/25

2.(-x-3)=53-27

2.(-x-3)=26

-x-3=26/2

-x-3=13

-x=13+3

-x=16

=> x=-16

2.(-x-3)+27=53

2.(-x-3)      =53-27

2.(-x-3)      =26

   (-x-3)      =26:2

   -x-3      =13

   => -x=  16

  => x= -16

2.(-x-3)+27=53

2.(-x-3)=53-27

2.(-x-3)=26

-x-3=26:2

-x-3=13

-x=13+3

-x=16

=>x=-16

Vậy x=-16

Bài 12: 

a. \(x< -\dfrac{170}{17}=-10\rightarrow\) x nguyên max = -11

b. \(x< -\dfrac{13}{3}=-4,\left(3\right)\rightarrow\)x nguyên max là -5

Bài 13: 

a. \(x>-\dfrac{12}{13}=-0,926..\rightarrow\)x nguyên min = 0 

b. \(x>-\dfrac{45}{15}=-3\rightarrow\)x nguyên min = -2

Cảm ơn mn

B12

a) ta có \(\dfrac{170}{17}=10\Rightarrow x=9\)

b) ta có  \(\dfrac{13}{3}=4,333...\Rightarrow x=4\)

B13

a) x>\(\dfrac{-12}{13}\Rightarrow x=-1\)

b) ta có: \(\dfrac{-45}{15}\Rightarrow x>-3\Rightarrow x=-2\)

\(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{11}{9}\)
\(x+\dfrac{3}{22}=\dfrac{3}{11}\)
\(x=\dfrac{3}{11}-\dfrac{3}{22}\)
\(x=\dfrac{3}{22}\)
#Đang Bận Thở

`x +3/22=27/121 .11/9`

`x+3/22=(27.11)/(121.9)`

`x +3/22=(9.3.11)/(11.11.9)`

`x+3/22=3/11`

`x = 3/11-3/22`

`x=6/22-3/22`

`x=3/22`

\(x+\dfrac{3}{22}=\dfrac{27}{121}.\dfrac{11}{9}\\ x+\dfrac{3}{22}=\dfrac{3}{11}\\x=\dfrac{3}{11}-\dfrac{3}{22}\\ x=\dfrac{3}{22} \)

a) x² - 16 - 4xy + 4y²

= ( x² - 4xy + 4y² ) - 16

= ( x - 2y )² - 4²

= ( x - 2y - 4 )( x - 2y + 4 )

b) x⁵ - x⁴ + x³ - x²

= x²( x³ - x² + x - 1 )

= x²[ x²( x - 1 ) + ( x - 1 ) ]

= x²( x - 1 )( x² + 1 )

c) x( x + 4 )( x + 6 )( x + 10 ) + 128 < mình nghĩ là nên sửa đề như này :]> 

= [ x( x + 10 ) ][ ( x + 4 )( x + 6 ) ] + 128

= ( x² + 10x )( x² + 10x + 24 ) + 128

Đặt t = x² + 10x

bthuc <=> t( t + 24 ) + 128

            = t² + 24t + 128

            = t² + 16t + 8t + 128

            = t( t + 16 ) + 8( t + 16 ) 

            = ( t + 16 )( t + 8 )

            = ( x² + 10x + 16 )( x² + 10x + 8 )

            = ( x² + 2x + 8x + 16 )( x² + 10x + 8 )

            = [ x( x + 2 ) + 8( x + 2 ) ]( x² + 10x + 8 )

            = ( x + 2 )( x + 8 )( x² + 10x + 8 )

cảm ơn bạn câu c mình chép nhầm nó là 128 đó 

= 3,45 x (123 + 4 - 27)

= 3,45 x 100

= 345

HT

345

34500