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\(1,=\left(x+1\right)\left(x^2+2x\right)=\left(x+1\right)x\left(x+2\right)\)
\(2,=x\left(2x-3\right)+2\left(2x-3\right)=\left(2x-3\right)\left(x+2\right)\)
\(3,=3x\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(3x-1\right)\)
\(4,=2x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(2x-6\right)=\left(x-y\right)2\left(x-3\right)\)
\(5,=4x\left(x+3\right)\)
BÀI 1:
a) \(x^4+2x^2y+y^2=\left(x^2+y\right)^2\)
b) \(\left(2a+b\right)^2-\left(2b+a\right)^2=\left(2a+b+2b+a\right)\left(2a+b-2b-a\right)\)
\(=\left(3a+3b\right)\left(a-b\right)=3\left(a+b\right)\left(a-b\right)\)
c) \(\left(a^3-b^3\right)+\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left[a^2+ab+b^2+\left(a-b\right)\right]=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
d) \(\left(x^2+1\right)^2-4x^2=\left(x^2+1-2x\right)\left(x^2+1+2x\right)=\left(x-1\right)^2\left(x+1\right)^2\)
e) \(\left(y^3+8\right)+\left(y^2-4\right)=\left(y+2\right)\left(y^2-y+2\right)\)
f) \(1-\left(x^2-2xy+y^2\right)=1-\left(x-y\right)^2=\left(1-x+y\right)\left(1+x-y\right)\)
g) \(x^4-1=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
h) ktra lại đề
m) \(\left(x-a\right)^4-\left(x+a\right)^4=-8ax\left(a^2+x^2\right)\)
a , \(-q^3+12q^2x-48qx^2+64x^3\)
\(=-\left(q^3-12q^2x+48qx^2-64x^3\right)\)
\(=\)\(-\left(q-4x\right)^3\)
b , x2 + 2xy - y2 - 9
= - ( x2 - 2xy + y2 ) - 9
= - ( x - y )2 - 9
= ( - x + y - 3 ) ( x - y + 3 )
3 , 1 - m2 + 2mn - n2
= 1 - ( m2 - 2mn + n2 )
= 1 - ( m - n )2
= ( 1 - m + n ) ( 1 + m - n )
4 , x3 - 8 + 6a2 - 12a
= x3 + 6a2 - 12a + 8
= x3 + 6a2 - 12a + 4 + 4
= x3 + ( 6a2 - 12a + 4 ) + 4
= x3 + ( 3a - 2 )2 + 4
= ( x + 3a - 2 + 2 ) ( x2 + 3a + 2 + 2 )
( Mai làm tiếp mấy ý sau '-' muộn rồi ~ )
5 , x2 - 2xy + y2 - xz - yz
= ( x2 - 2xy + y2 ) - ( xz + yz )
= ( x - y )2 - z ( x + y )
= ( x - y ) 2 - z ( x - y )
= ( x - y ) ( x - y - z )
6 , x2 - 4xy + 4y 2 - z2 + 4z - 4t2
=( x2 - 4xy + 4y 2 ) - (z2 - 4z +4 ) . t2
= ( x - y )2 - ( z - 2 )2 . t2
= ( x - y - z - 2 ) ( x - y + z - 2 ) t2
7 , 25 - 4x2 - 4xy - y2
= 25 + ( - 4x2 - 4xy + y2 )
= 25 + ( 2x - y )2
= ( 5 + 2x - y ) ( 5 + 2x + y )
8 ,
x3 + y3 + z3 - 3xyz
= (x+y)3 - 3xy (x - y ) + z3 - 3xyz
= [ ( x + y)3 + z3 ] - 3xy ( x + y + z )
= ( x + y + z )3 - 3z ( x + y )( x + y + z ) - 3xy ( x - y - z )
= ( x + y + z )[( x + y + z )2 - 3z ( x + y ) - 3xy ]
= ( x + y + z )( x2 + y2 + z2 + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= ( x + y + z)(x2 + y2 + z2 - xy - xz - yz)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
Trả lời:
1) sửa đề: \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)
2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)
3) \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)
a/ \(x^3=5x-12\Leftrightarrow x^3-5x+12=0\Leftrightarrow\left(x^3+3x^2\right)-\left(3x^2+9x\right)+\left(4x+12\right)=0\)
\(\Leftrightarrow x^2\left(x+3\right)-3x\left(x+3\right)+4\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+4\right)=0\)
*) x + 3 = 0 <=> x = -3
S = {-3}
b/ có ng giải
c/ \(\left(2x^2-5x+3\right)^2=\left(x^2+x-2\right)^2\Leftrightarrow\left(2x^2-5x+3\right)^2-\left(x^2+x-2\right)^2=0\)
\(\Leftrightarrow\left(2x^2-5x+3-x^2-x+2\right)\left(2x^2-5x+3+x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-6x+5\right)\left(3x^2-4x-1\right)=0\)
\(\Leftrightarrow\left[\left(x^2-x\right)-\left(5x+5\right)\right]\left(3x^2-4x+1\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-5\left(x-1\right)\right]\left(3x^2-4x+1\right)=0\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(3x^2-4x+1\right)=0\)
*) x- 5 = 0 <=> x = 5
*) x- 1 = 0 <=> x = 1
S={1;5}
d/ \(x^3-x^2=4\left(x-1\right)^2\Leftrightarrow x^3-x^2-4\left(x-1\right)^2=x^3-x^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-5x^2+8x-4=\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2=0\)
*) x - 1 = 0 <=> x = -1
*) (x - 2)^2 = 0 <=> x = 2
S = {-1;2}
1, \(x^2+4x-2xy-4y+y^2=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)=\left(x-y\right)^2+4\left(x-y\right)=\left(x-y\right)\left(x-y+4\right)\)
2, \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
3, \(2x^2+4x+2-2y^2=2\left(x^2-y^2\right)+2\left(2x+1\right)=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x+1-y\right)\left(x+1+y\right)\)
4, \(x^4-2x^2=x^4-2x^2+1-1=\left(x^2-1\right)^2-1=\left(x^2-1-1\right)\left(x^2-1+1\right)=\left(x^2-2\right)x^2\)
5, \(x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)=x\left[\left(x+y\right)^2-3^2\right]=x\left(x+y-3\right)\left(x+y+3\right)\)
6, \(x^3-\frac{1}{4}x=x\left(x^2-\frac{1}{4}\right)=x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)
7, \(2x-2y-x^2+2xy-y^2=\left(2x-2y\right)-\left(x^2-2xy+y^2\right)=2\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(2-x+y\right)\)
8, \(\left(2x+3\right)^2-\left(x+1\right)^2=\left(2x+3+x+1\right)\left(2x+3-x-1\right)=\left(3x+4\right)\left(x+2\right)\)
a)x4+2x3+5x2+4x-12
=(x4+2x3+x2)+(4x2+4x)-12
=(x2+x)2+4(x2+x)-12
Đặt t=x2+x
=t2+4t-12=(t-2)(t+6)
=(x2+x-2)(x2+x+6)
=(x-1)(x+2)(x2+x+6)
b)(x+1)(x+2)(x+3)(x+4)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt x2+5x+4=t
t(t+2)+1=t2+2t+1
=(t+1)2=(x2+5x+4+1)2
=(x2+5x+5)2
c)(x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+5)
=(x2+8x+10)(x+2)(x+6)
d)(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4
t(t+2)-24=(t-4)(t+6)
=(x2+5x+4-4)(x2+5x+4+6)
=x(x+5)(x2+5x+10)
a) \(2x^2-7xy+5y^2=2x^2-2xy-5xy+5y^2=2x\left(x-y\right)-5y\left(x-y\right)=\left(x-y\right)\left(2x-5y\right)\)
b) \(5x^3+10x^2y+5xy^2=5x\left(x^2+2xy+y^2\right)=5x\left(x+y\right)^2\)
c) \(x^2-2xy+y^2-9=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
d) \(x\left(x-2\right)+x-2=\left(x-2\right)\left(x+1\right)\)
e) \(5x\left(x-3\right)-x+3=\left(x-3\right)\left(5x-1\right)\)