Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-\frac{9}{4}\right)\left(x+\frac{4}{3}\right)\left(120x^3+12x^2-24x+36\right)\)
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)
\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1-2x\right)=\left(x^2-1\right)\left(x-1\right)^2\)
\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27\)
\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)
\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)
\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)
\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)
1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)
Bạn tự lm tiếp.AD HĐT số (7)
2)\(\left(x+4\right)^3-64=\left(x+4\right)^3-4^3\)
AD HĐT số (7).Tự lm tiếp
3)\(x^6+1=\left(x^2\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
4)\(x^9+1=\left(x^3\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)
AD HĐT số (3).Tự lm tiếp
6)\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
AD HĐT số (4)
7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)
AD HĐT số (5)
8)\(27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
AD HĐT số (5)
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)