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a.5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[ (x2-2xy+y2)-(2z)2 ]
=5[ (x-y)2-(2z)2 ]
=5(x-y-2z)(x-y+2z)
b.16x-5x2-3
=15x+x-5x2-3
=(15x-3)+(x-5x2)
=3(5x-1)+x(1-5x)
=3(5x-1)-x(5x-1)
=(5x-1)(3-x)
c.x2-5x+5y-y2
=(5y-5x)+(x2-y2)
=5(y-x)+(x-y)(x+y)
=5(y-x)-(y-x)(y+x)
=(y-x)[5-(y+x)]
=(y-x)(5-y-x)
d.3x2-6xy+3y2-12z2 (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)
=3(x2-2xy+y2-4z2)
=3[ (x2-2xy+y2)-(2z)2 ]
=3[ (x-y)2-(2z)2 ]
=3(x-y-2z)(x-y+2z)
e.x2+4x+3
=x2+3x+x+3
=(x2+x)+(3x+3)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f.(x2+1)2-4x2
=(x2+1)2-(2x)2
=(x2+1-2x)(x2+1+2x)
h.x2-4x-5
=x2-5x+x-5
=(x2+x)+(-5x-5)
=x(x+1)-5(x+1)
-(x+1)(x-5)
\(1.5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x^2-2xy+y^2\right)-\left(2z\right)^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
\(2.16x-5x^2-3\)
\(=-\left(5x^2-16x+3\right)\)
\(=-\left(5x^2-15x-x+3\right)\)
\(=-\left[\left(5x^2-15x\right)-\left(x-3\right)\right]\)
\(=-\left[5x\left(x-3\right)-\left(x-3\right)\right]\)
\(=-\left(x-3\right)\left(5x-1\right)\)
\(3.x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-\left(5x-5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
\(4.3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
\(5.x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=\left(x^2+3x\right)+\left(x+3\right)\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x+1\right)\)
\(6.\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
\(7.x^2-4x-5\)
\(=x^2-5x+x-5\)
\(=\left(x^2-5x\right)-\left(x-5\right)\)
\(=x\left(x-5\right)-\left(x-5\right)\)
\(=\left(x-5\right)\left(x-1\right)\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
5x2 - 10xy + 5y2 - 20z2
= 5.(x2 - 2xy + y2 - 4z2)
= 5.[(x2 - 2xy + y2) - (2z)2]
= 5.[(x - y)2 - (2z)2]
= 5.(x - y - 2z).(x - y + 2z)
x2.(1 - x2) - 4 + 4x2
= x2.(1 - x2) - 4.(1 - x2)
= (1 - x2).(x2 - 4)
= (1 - x)(1 + x)(x - 2)(x + 2)
5x2 - 10xy + 5y2 - 20z2
= 5.(x2 - 2xy + y2 - 4z2)
= 5.[(x2 - 2xy + y2) - (2z)2]
= 5.[(x - y)2 - (2z)2]
= 5.(x - y - 2z).(x - y + 2z)
x2.(1 - x2) - 4 + 4x2
= x2.(1 - x2) - 4.(1 - x2)
= (1 - x2).(x2 - 4)
= (1 - x)(1 + x)(x - 2)(x + 2)
a )
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-4z^2\right]\)
b )
\(-5x^2-16x-3\)
\(=-5x^2-15x-x-3\)
\(=-5x\left(x+3\right)-\left(x+3\right)\)
\(=\left(-5x-1\right)\left(x+3\right)\)
c )
\(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)-5\right]\)
d )
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
P/s : Mình bổ sung :
a )
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
d )
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
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)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3.\left[\left(x+y\right)^2-z^2\right]=3.\left(x+y-z\right)\left(x+y+z\right)\)
\(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
a, x^2-9+(x-3)^2 = (x-3)(x+3)+(x-3)^2=(x-3)(x+3+x-3)=2x(x-3)
b,có sai k ạ ! vì mình thấy tự nhiên có ẩn y ở đó , nếu đề bài 2 ẩn thì 1 trong 3 hạng tử chứa ẩn x kia phải có thêm 1 ẩn y
c,đề bài thiếu ẩn ở hạng tử thứ nhất ạ !
a) \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
b) \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
g) \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
a,\(x^3-4x^2-9x+36\)
\(=\left(x^3-4x^2\right)+\left(-9x+36\right)\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x^2-9\right)\left(x-4\right)\)