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a)
\(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
b)
\(=a\left(a-b\right)+a-b\)
\(=\left(a+1\right)\left(a-b\right)\)
c)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left(x+1-y\right)\left(x+1+y\right)\)
d)
\(=x^3\left(x-2\right)+10x\left(x-2\right)\)
\(=x\left(x^2+10\right)\left(x-2\right)\)
e)
\(=x\left(x^2+2x+1\right)\)
\(=x\left(x+1\right)^2\)
f)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
a,2x3+3x2+2x+3
=(2x3+2x)+(3x2+3)
=2x(x2+1)+3(x2+1)
=(x2+1)(2x+3)
b,a2-ab+a-b
=(a2-ab)+(a-b)
=a(a-b)+(a-b)
=(a-b)(a+1)
c,2x2+4x+2-2y2
=2(x2+2x+1-y2)
=2[(x2+2x+1)-y2 ]
=2[(x+1)2-y2 ]
=2(x+1-y)(x+1+y)
d,x4-2x3+10x2-20x
=(x4-2x3)+(10x2-20x)
=x3(x-2)+10x(x-2)
=(x-2)(x3+10x)
=(x-2)[x(x2+10)]
e,x3+2x2+x
=x(x2+2x+1)
=x(x+1)2
f,xy+y2-x-y
=(xy+y2)-(x-y)
=y(x+y)-(x+y)
=(x+y)(y-1)
a,\(x^2y^2+y^3+zx^2+yz=\left(x^2y^2+y^3\right)+\left(zx^2+yz\right)\)
\(=y^2\left(x^2+y\right)+z\left(x^2+y\right)\)
\(=\left(y^2+z\right)\left(x^2+y\right)\)
b,\(x^4+2x^3-4x-4=x^4+2x^3+x^2-x^2-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
c,\(x^3+2x^2y-x-2y=\left(x^3+2x^2y\right)-\left(x+2y\right)\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x^2-1\right)\left(x+2y\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2y\right)\)
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
a) x3 - x2 - 5x + 125
=(x3-6x2+25x)+(5x2-30x+125)
=x(x2-6x+25)+5(x2-6x+25)
=(x+5)(x2-6x+25)
b) x3 + 2x2 - 6x - 27
=x3+5x2+9-3x2-15x-27
=x(x2+5x+9)-3(x2+5x+9)
=(x-3)(x2+5x+9)
c) 12x3 + 4x2 - 27x - 9
=4x2(3x+1)-9(3x+1)
=(4x2-9)(3x+1)
=[(2x)2-32](3x+1)
=(2x-3)(2x+3)(3x+1)
a) \(x^3-x^2-5x+125\)
\(=\left(x^3+125\right)-\left(x^2+5x\right)\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(x^3+2x^2-6x-27\)
\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+9+2x\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
c) \(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x-1\right)\left(4x^2-9\right)=\left(3x-1\right)\left(2x-3\right)\left(2x+3\right)\)
\(a)\) \(x^2-2x-4y^2-4y\)
\(=\)\(\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\)\(\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\)\(\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\)\(\left(x-2y-2\right)\left(x+2y\right)\)
\(=\)\(2\left(x-y\right)\left(x+2y\right)\)
Chúc bạn học tốt ~
a) Ta có x2 - 2x - 4y2 - 4y
= x2 - 2x + 1 - 4y2 - 4y - 1
= (x - 1)2 - (4y2 + 4y + 1)
= (x - 1)2 - (2y + 1)2
= (x - 1 - 2y - 1)(x - 1 + 2y + 1)
= (x - 2y - 1)(x + 2y)
Bài 1:(Theo mình câu a nên sửa lại như thế này nhé)
a, a2-5a-14 b,x4+x2-2
=a2+2a-7a-14 =x4-x3+x3-x2+2x2-2x+2x-2
=(a2+2a)-(7a+14) =(x4-x3)+(x3-x2)+(2x2-2x)+(2x-2)
=a(a+2)-7(a+2) =x3(x-1)+x2(x-1)+2x(x-1)+2(x-1)
=(a+2)(a-7) =(x-1)(x3+x2+2x+2)
=(x-1)[(x3+x2)+(2x+2)]
=(x-1)[x2(x+1)+2(x+1)]
=(x-1)(x+1)(x2+2)
Bài 2:
a, x3+x2+x+1=0
<=>(x3+x2)+(x+1)=0
<=>x2(x+1)+(x+1)=0
<=>(x+1)(x2+1)=0
<=>\(\orbr{\begin{cases}x+1=0\\x^2+1=0\left(loại\right)\end{cases}}\)(x2 luôn lớn hơn hoặc bằng 0 =>x2+1 luôn lớn hơn hoặc bằng 1 nên x2+1=0 loại nhé)
<=>x= -1
b, x(2x-7)-4x+14=0
<=>x(2x-7)-(4x-14)=0
<=>x(2x-7)-2(2x-7)=0
<=>(2x-7)(x-2)=0
<=>\(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
a/
\(x^3\left(x-4\right)+x\left(x-4\right)=\left(x-4\right)\left(x^3+x\right)=\)
\(=x\left(x^2+1\right)\left(x-4\right)\)
b/
\(\left(x-1\right)\left(x^2-9\right)-2x\left(x-3\right)=\)
\(=\left(x-1\right)\left(x-3\right)\left(x+3\right)-2x\left(x-3\right)=\)
\(=\left(x-3\right)\left[\left(x-1\right)\left(x+3\right)-2x\right]=\)
\(=\left(x-3\right)\left(x^2-3\right)\)