Question

Câu 1:

a,Làm phép tính nhân:(x-2)(x+5)

b,Phân tích đa thức thành nhân tử:3x+3y+ax+ay

c,Làm phép tính chia:(x^2+2xy):(x+2y)

d,Tính sao cho:(x-2)(x+2)+(x+1)^2-2x^2=0

Làm giúp mk vs 

thanhks

Answer 1

a ) ( x - 2 )( x + 5 )

= x^2 + 5x - 2x + 10

= x^2 + 3x + 10

b ) 3x + 3y +ax + ay 

= x( 3 + a ) + y( 3 + a )

= ( 3 + a )( x + y ) 

c ) ( x^2 + 2xy ) : ( x + 2y )

= [ x( x + 2y ) ] : ( x + 2y )

= x  : 1

= x 

d ) ( x - 2 )( x + 2 ) + ( x + 1 )^2 - 2x^2 = 0 

       x^2 + 2x - 2x - 4 + x^2 + x + x + 1 - 2x^2 = 0

       x^2 - 4 + x^2 + 2x + 1 - 2x^2 = 0

        2x^2 + 2x - 4 + 1 - 2x^2 = 0 

       2x - 3 = 0

       2x = 0 + 3

       2x = 3

        x = 3 : 2

        x = 3/2 

Answer 2

a)   \(\left(x-2\right)\left(x+5\right)\)

\(=x^2+5x-2x-10\)

\(=x^2+3x-10\)

b)  \(3x+3y+ax+ay\)

\(=3\left(x+y\right)+a\left(x+y\right)\)

\(=\left(x+y\right)\left(3+a\right)\)

c)   \(\left(x^2+2xy\right):\left(x+2y\right)\)

\(=\left[x\left(x+2y\right)\right]:\left(x+2y\right)\)

\(=x\)

d)  \(\left(x-2\right)\left(x+2\right)+\left(x+1\right)^2-2x^2=0\)

\(\Leftrightarrow\)\(x^2-4+x^2+2x+1-2x^2=0\)

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

\(\Leftrightarrow\)\(2x=3\)

\(\Leftrightarrow\)\(x=\frac{3}{2}\)

Vậy....

Answer 3

(x-2)(x+5)=x^2+5x-2x-10=x^2+3x-10

3x+3y+2x+2y=3(x+y)+2(x+y)=5(x+y)

(x^2+2xy):(x+2y)=x(x+2y):(x+2y)=x

x^2-4+x^2+2x+1-2x^2=0 khi 2x-3=0 khi x=3/2