cái này có những biểu thức ko dùng được hđt hoặc đặt nhân tử chung =))) anh nghĩ đề em nên thay chữ và thành hoặc 

1, \(15x^2+10xy=5x\left(3x+2y\right)\)

2, \(24x-18y+30=6\left(4x-3y+5\right)\)

3, \(2x\left(y-2009\right)+5y\left(y-2009\right)=\left(y-2009\right)\left(2x+5y\right)\)

4, \(35x\left(y-8\right)-14y\left(8-y\right)=\left(y-8\right)\left(35x+14y\right)=7\left(5x+2y\right)\left(y-8\right)\)

5, \(x^2+14x+49=x^2+2.7x+7^2=\left(x+7\right)^2\)

6, \(9x^2-4=\left(3x-2\right)\left(3x+2\right)\)

Trả lời:

1, 15x² + 10xy = 5x ( 3x + 2y )

2, 24x - 18y + 30 = 6 ( 4x - 3y + 5 )

3, 2x ( y - 2009 ) + 5y ( y - 2009 ) = ( y - 2009 )( 2x + 5y )

4, 35x ( y - 8 ) - 14y ( 8 - y ) = 35x ( y - 8 ) + 14y ( y - 8 ) = ( y - 8 )( 35x + 14y ) = 7 ( y - 8 )( 5x + 2y )

5, x² + 14x + 49 = ( x + 7 )² 

6, 9x² - 4 = ( 3x - 2 )( 3x + 2 )

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

Dễ nhưng mà dài chết người 

Bài 1:

Ta có: \(x^2-2x+2=x^2-2x+1+1\)

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

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

Ta thấy rằng: \(\left(x-1\right)^2\ge0\) ( Với mọi \(x\in Z\) )

mà 1 > 0

=> \(\left(x-1\right)^2+1\ge0\)

<=> \(x^2-2x+1\ge0\)

Bài 3:

a) 53^2 + 47^2 + 94.53

= 53^2 + 47^2 + 2.47.53

= ( 53 + 47 )^2

= 100^2

= 10000

b) 50^2 - 49^2 + 48^2 - 47^2 + 2^2 - 1^2

= ( 50^2 - 49^2 ) + ( 48^2 - 47^2 ) + ( 2^2 - 1^2 )

= (50+49).(50-49) + (48+47).(48-47) + (2+1).(2-1)

= 50 + 49 + 48 + 47 + 2 + 1

= (49 + 1) + (48 + 2) + 50 + 47

= 50 + 50 + 50 + 47

= 197

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)

5x^2+5y^2+8xy-2x+2y+2=0

=>(4x^2+8xy+4y^2)+(x^2-2x+1)+(y^2+2y+1)=0

=>(2x+2y)^2+(x-1)^2+(y+1)^2=0

tổng 3 biểu thức không âm = 0 <=> chúng đều = 0

<=>2(x+y)=x-1=y+1=0

=>x=1;y=-1

Thay vào M ........

thanks hoàng phúc nha!

2x^2-5x-7

=2x^2+2x-7x-7

=2x.(x+!) -7.(x+1)

(2x-7).(x+1)

x^2-y^2-5x+5y

=(x-y).(x+y)-5.(x-y)

=(x-y-5).(x+y)

5x^3-5x^2.y-10x^2+10xy

=5x^2.(x-y)-10x.(x-y)

=(5x^2-10x).(x-y)

=5x.(x-2).(x-y)

a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)

b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)

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

c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)

\(=\left(3x-5\right)\left(x+1\right)\)

a) x⁴ + 2x³ + x²

= x² ( x² + 2x + 1 )

= x² ( x + 1 )² 

b) 5x² - 10xy + 5y² - 20z² 

= 5 [(x² - 2xy + y² ) - 4z² ] 

= 5 [( x - y )² - ( 2z )² ]

= 5 ( x - y - 2z ) ( x - y + 2z )

 c) x³ - x + 3x²y + 3xy²+ y³- y

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= (x + y )³ - ( x + y)

= ( x + y ) [( x + y )² - 1 ]

= ( x + y ) ( x + y + 1 ) ( x + y - 1 )