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a) ( x-3y ) ( x + 1 )
b) ( x+y+5 ) ( x+y-5 )
c) ( x-5 ) ( x+2 )
Hk tốt
a)\(x^2-6xy+9y^2=x^2-2\cdot x\cdot3y+\left(3y\right)^2=\left(x-3y\right)^2\)
b) \(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2+1+2x\right)\left(x^2+1-2x\right)\)
\(=\left(x+1\right)^2\left(x-1\right)^2\)
a)\(x^2-6xy+9y^2-25z^2=\left[x^2-2.x.3y+\left(3y\right)^2\right]-\left(5z\right)^2\)
\(=\left(x-3y\right)^2-\left(5z\right)^2=\left(x-3y-5z\right)\left(x-3y+5z\right)\)
b)\(xyz+x^2yz-6yz=yz\left(x^2+x-6\right)=yz\left(x^2+3x-2x-6\right)\)
\(=yz\left[x\left(x+3\right)-2\left(x+3\right)\right]=yz\left(x-2\right)\left(x+3\right)\)
a.x2-6xy+9y2-25z2
= ( x2-6xy+9y2)-25z2
= [x2-2x3y+(3y)2]-25z2
= (x-3y)2-252
= (x-3y+25)(x-3y-25)
a) 2x3 + 6xy - x2z - 3yz
= ( 2x3 + 6xy ) - ( x2z + 3yz )
= 2x( x2 + 3y ) - z( x2 + 3y )
= ( x2 + 3y )( 2x - z )
b) x2 - 6xy + 9y2 - 49
= ( x2 - 6xy + 9y2 ) - 49
= ( x - 3y )2 - 72
= ( x - 3y - 7 )( x - 3y + 7 )
c) x3 + 4x2 + 16x + 64
= ( x3 + 4x2 ) + ( 16x + 64 )
= x2( x + 4 ) + 16( x + 4 )
= ( x + 4 )( x2 + 16 )
a) =(2x^3-x^2z)+(6xy-3yz)
=x^2(2x-z)+3y(2x-z)
=(x^2+3y)(2x-z)
b) =(x^2-6xy+9y^2)-7^2
=(x-3y)^2-7^2
=(x-3y+7)(x-3y-7)
c) =(x^3+4x^2)+(16x+64)
=x^2(x+4)+16(x+4)
=(x^2+16)(x+4)
a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
a) A = (x + 1)(y - 2) - (2 - y)2
= -[(x + 1)(2 - y) + (2 - y)2]
= -[(x + 1 - 2 + y)(2 - y)]
= -[(x - 1 + y)(2 - y)]
= (x - 1 + y)(y - 2)
Bài 2:
a) \(A=\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)
\(A=\left(x+1\right)\left(y-2\right)-\left(y-2\right)^2\)
\(A=\left(y-2\right)\left(x+1-y+2\right)\)
\(A=\left(y-2\right)\left(x-y+3\right)\)
b) \(B=x^2-6xy+9y^2+4x-12y\)
\(B=\left[x^2-2\cdot x\cdot3y+\left(3y\right)^2\right]+4\left(x-3y\right)\)
\(B=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(B=\left(x-3y\right)\left(x-3y+4\right)\)
Bài 3:
a) \(3\left(x-2\right)\left(x+3\right)-x\left(3x+1\right)=2\)
\(\left(3x^2+3x-18\right)-\left(3x^2+x\right)-2=0\)
\(3x^2+3x-18-3x^2-x-2=0\)
\(2x-20=0\)
\(x=10\)
b) \(6x^2+13x+5=0\)
\(6x^2+10x+3x+5=0\)
\(2x\left(3x+5\right)+\left(3x+5\right)=0\)
\(\left(3x+5\right)\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+5=0\\2x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{-1}{2}\end{cases}}}\)
\(x^2+4x-9y^2+4\)
\(=\left(x^2+2.2x+2^2\right)-\left(3y\right)^2\)
\(=\left(x+2\right)^2-\left(3y\right)^2\)
\(=\left(x+2-3y\right)\left(x+2+3y\right)\)
\(x^2-9y^2-6y-1\)
\(=x^2-\left[\left(3y\right)^2+2.3y+1^2\right]\)
\(=x^2-\left(3y+1\right)^2\)
\(=\left(x-3y-1\right)\left(x+3y+1\right)\)
Tham khảo nhé~
a/Ta có:x2+4x-9y2+4
=x2+4x+4-(3y)2
=(x+2)2-(3y)2
=(x+2-3y)(x+2+3y)
b/Ta có:x2-9y2-6xy-1
=x2-6xy-(3y)2-1
=(x-3y-1)(x-3y+1)
a) 7(x-y)
b) (x-3y)2
c) xy(x+y)-(x+y)=(xy-1)(x+y)
d) x2-3x+2x-6=x(x-3)+2(x-3)=(x+2)(x-3)
x^2+6xy+9y^2-3x-9y+2
=( x^2+6xy+9y^2)-3(x+3y)+9/4 -1/4
=(x+3y)^2-3(x+3y)+(3/2)^2- 1/4
=(x+3y+3/2)^2-(1/2)^2
=(x+3y+3/2+1/2)(x+3y+3/2-1/2)=(x+3y+2)(x+3y+1)