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) 2x(y-2009)+5y(y-2009)

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

b) 35x(y-8)-14y(8-y)

=35x(y-8)+14y(y-8)

=7(y-8)(5x+2y)

c) 15x^2 +10xy =5x(3x+2y)

d)24x-18y+30= 6(4x+3y+10)

e) x^2 +14x+49

= x^2 +2.7.x+ 7^2

=(x+7)^2

i) x^2 -x - y^2 - y

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

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

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

k) x^2 -2xy+y^2 -z^2

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

=(x-y)^2 -z^2

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

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

b) \(35x(y-8)-14y(8-y)\) \(=35x\left(y-8\right)+14y\left(y-8\right)\)

\(=\left(y-8\right)\left(35x+14y\right)\)

\(=\left(y-8\right).7\left(5x+2y\right)\)

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

d) \(24x-18y+30=3\left(8x-6y+10\right)\)

e) \(x^2+14x+49=x^2+2.7.x+7^2\)

\(=\) \((x+7)^2\)

g) \(27x^3+y^3\) \(=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)

h) \(8x^3-\dfrac{1}{125}y^3=\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

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

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

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

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

\(=\left(x-y\right)^2-z^2\)

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

- HỌC TỐT NHA BẠN YÊU ^^ -

1) Ta có: \(x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

1) \(5x-5y+x\left(x-y\right)\)

\(=5\left(x-y\right)+x\left(x-y\right)\)

\(=\left(x-y\right)\left(x+5\right)\)

2) \(x^2+4x+3\)

\(=\left(x^2+x\right)+\left(3x+3\right)\)

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

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

3) \(x^2-2xy+y^2-z^2\)

\(=\left(x-y\right)^2-z^2\)

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

4) \(x\left(x-5\right)-3x+15\)

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

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

5) \(y^2-x^2+2x-1\)

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

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

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

\(1.\left(x-y\right)\left(x+5\right)\)

\(2.\left(x+1\right)\left(x+3\right)\)

\(3.\left(x-y-z\right)\left(x-y+z\right)\)

\(4.\left(x-3\right)\left(x-5\right)\)

\(5.\left(y-x+1\right)\left(y+x+1\right)\)

\(7.\left(x+1\right)\left(x-2\right)^2\)

\(8.\left(x-5\right)\left(x+3\right)\)

\(10.\left(y+1\right)\left(2x+z\right)\)

1)

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

2)

x²+4x+3=x²+x+3x+3=(x+1)(x+3)

3)x²-2xy+y²-z²=\(\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

4)\(x\left(x-5\right)-3x+15=\left(x-3\right)\left(x-5\right)\)

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!

a. y⁴ - 14y² + 49

Gọi y² là t, ta có:

t² - 14t + 49

<=> t² - 14t + 7²

<=> (t - 7)²

Thay x² = t

<=> (x² - 7)²

b. \(\dfrac{1}{4}-x^2\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)

\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

c. x⁴ - 16

<=> (x²)² - 4²

<=> (x² - 4)(x² + 4)

d. x² - 9

<=> x² - 3²

<=> (x - 3)(x + 3)

= ( 8 - y )( 35x - 14y )