Trả lời :

5 . (x - 1) + 10 = 0

<=> 5x - 5 + 10 = 0

<=> 5x = 5

<=> x = 1

Vậy x = 1 thì biểu thức 5 . (x - 1) + 10 có giá trị bằng 0.

Mình làm nhầm ạ, làm lại :

5 . (x - 1) + 10 = 0

<=> 5x - 5 + 10 = 0

<=> 5x + 5 = 0

<=> x = - 1

Vậy x = - 1 thì biểu thức 5 . (x - 1) + 10 có giá trị bằng 0.

They fly kites.

They play hide and seek.

They go swimming in the ponds.

They play outdoors with things they found.

5x(3x²-4x-1)=15x³-20x²-5x

a) 15x² + 10x = 5x ( 3x +2 )

b) x² - y² +2x -2y

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

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

c) 3x² + 6xy + 3y² -12 z²

= 3 ( x² +2xy+y²) -12z²

= 3 (x+y)²  - 12 z²

= 3 [(x+y)² - 4z² ]

= 3( x + y - 2z )( x + y + 2z)

d) x² - 5 -6 = x² -11

= (x - √11)(x+√11)

a, \(15x^2+10x=5x\left(3x+2\right)\)

b, \(x^2-y^2+2x-2y=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+2\right)\)

c, \(3x^2+6xy+3y^2-12z^2\)

\(=3\left(x^2+2xy+y^2-4z^2\right)\)

\(=3\left[\left(x+y\right)^2-\left(2z\right)^2\right]=3\left(x+y-2z\right)\left(x+y+2z\right)\)

d, \(x^2-5x-6=x^2-6x+x-6=x\left(x+1\right)-6\left(x+1\right)=\left(x-6\right)\left(x+1\right)\)

\(x^2+y^2-x^2y^2+xy-x-y\)

\(=x^2-x^2y^2+y^2-y+xy-x\)

\(=x^2\left(1-y^2\right)+y\left(y-1\right)+x\left(y-1\right)\)

\(=x^2\left(1-y\right)\left(y+1\right)+y\left(y-1\right)+x\left(y-1\right)\)

\(=\left(y-1\right)\left[-x^2\left(y+1\right)+y-x\right]\)

\(=\left(y-1\right)\left[-x^2y-x^2+y-x\right]\)