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\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
1)2xy+3z+6y+xz
= x(2y + z) + 3(z + 2y)
= (x + 3)(2y + z)
2)x^4-9x^3+x^2-9x
= x^2(x^2 + 1) - 9x(x^2 + 1)
= (x^2 + 1)(x^2 - 9x)
= x(x - 9)(x^2 + 1)
3)x^2-xy+x-y
= x(x - y) + (x - y)
= (x + 1)(x - y)
4)xz+yz-5(x+y)
= z(x + y) - 5(x + y)
= (z - 5)(x + y)
5)3x^2-3xy-5x+5y
= 3x(x - y) - 5(x - y)
= (3x - 5)(x - y)
6)x^2+4x-y^2+4y
= (x - y)(x + y) + 4(x + y)
= (x - y + 4)(x + y)
\(10x\left(x-y\right)-6y\left(y-x\right)\)
\(=10x\left(x-y\right)+6x\left(x-y\right)\)
\(=\left(10x+6x\right)\left(x-y\right)\)
\(c,3x^2+5y-3xy-5x\)
\(=\left(3x^2-3xy\right)+\left(5y-5x\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(3x-5\right)\left(x-y\right)\)
\(e,27+27x+9x^2=3\left(9+9x+x^2\right)\)
b)x2+2xy+y2-16=(x+y)2-42=(x+y+4)(x+y-4)
c)3x2+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)
d)4x2-6x3y-2x2+8x=2x(2x-3x2y-x+4)
e)x2-4-2xy+y2=(x2-2xy+y2)-4=(x-y)2-22=(x-y-2)(x-y+2)
k)x2-y2-z2-2yz=x2-(y+z)2=(x-y-z)(x+y+z)
m)6xy+5x-5y-3x2-3y2=3(x2-2xy+y2)+5(x-y)=3(x-y)2+5(x-y)=(x-y)(3x-3y+5)
a) \(3x^2-6x+9x^2\)
\(=12x^2-6x\)
\(=6x\left(2x-1\right)\)
b) hỏi chấm đề ?
c) \(3x^2+5y-3xy-5x\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
d) \(3y^2-3z^2+3x^2-6xy\)
\(=3\left(y^2-z^2+x^2-2xy\right)\)
\(=3\left[\left(x-y\right)^2-z^2\right]\)
\(=3\left(x-y-z\right)\left(x-y+z\right)\)
e) \(16x^3+54y^3\)
\(=2\left(8x^3+27y^3\right)\)
\(=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
f) \(x^2-25-2xy+y^2\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
g) \(x^5-3x^4-3x^3-x^2\)
\(=x^2\left(x^3-3x^2-3x-1\right)\)
1/x^3 - 2x^2 - 9x + 18
= x\(^2\)( x - 2 ) - 9 ( x - 2 ) = ( x\(^2\) - 9 ) ( x - 2 )= ( x - 3 ) ( x +3 ) ( x - 2 )
2/3x^2 -5x - 3y^2 + 5y
= 3( x\(^2\) - y\(^2\) ) - 5 ( x - y ) = 3 ( x - y ) ( x + y ) - 5 ( x - y ) = ( x - y ) [ 3( x+ y ) - 5 ]
= ( x - y ) ( 3x + 3y - 5 )
3/49 - x^2 + 2xy - y^2
= 49 - ( x\(^2\) - 2xy + y\(^2\) ) = 49 - ( x - y )\(^2\) = ( 7 - x + y ) ( 7 + x - y )
5/ x^2 - 4x^2y^2 + 2xy
= x ( x - 4xy\(^2\) + 2y )
6/ 3x - 3y - x^2 + 2xy - y^2
= ( 3x - 3y ) - ( x\(^2\) - 2xy + y\(^2\) ) = 3 ( x - y ) - ( x - y )\(^2\) = ( x - y ) ( 3 - x + y )
1)
a) => 16x2 - 8x + 1 - 8(2x2 + 3x - 4x - 6) = 15
=> 16x2 - 8x + 1 - 8(2x2 - x - 6) = 15
=> 16x2 - 8x + 1 - 16x2 + 8x + 48 = 15
=> 49 = 15 (?) (vô lí)
=> Không tìm được x thoả mãn
b) (5x - 2)(x - 2) - 4(x - 3) = x2 + 3
=> 5x2 - 10x - 2x + 4 - 4x + 12 = x2 + 3
=> 5x2 - 16x + 16 = x2 + 3
=> 4x2 - 16x + 16 = 3
=> (2x)2 - 2.2x.4 + 42 = 3
=> (2x - 4)2 = 3
=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)
Mong bạn xem lại đề bài!
2)
a) 5x2 - 10xy + 5y2 - 20z2
= 5(x2 - 2xy + y2 - 4z2)
= 5[(x - y)2 - (2z)2]
= 5(x - y - 2z)(x - y + 2z)
b) a3 - ay - a2x + xy
= a(a2 - y) - x(a2 - y)
= (a - x)(a2 - y)
c) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x - y)2 - (2z)2]
= 3(x - y - 2z)(x - y + 2z)
d) x2 - 2xy + tx - 2ty
= x(x - 2y) + t(x - 2y)
= (x + t)(x - 2y)
e: \(49-x^2-2xy-y^2\)
\(=49-\left(x+y\right)^2\)
\(=\left(7-x-y\right)\left(7+x+y\right)\)
d: \(16x^2-24xy+9y^2=\left(4x-3y\right)^2\)
Làm cho em thêm mấy câu trên với ạ