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a) (x+2)(x-3)=0
<=> x+2=0
x-3=0
<=> x=-2
x= 3
b) 2x-x2=0
<=> x(2-x) =0
<=> x=0
2-x=0
<=> x=0
x=2
a)(x+2)(x-3)=0
=>\(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)=>\(\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
Vậy x=-2 hoặc x=3
b) 2x-x2=0
=> x(2-x)=0
=>\(\orbr{\begin{cases}x=0\\2-x=0\end{cases}}\)=>\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy x=0 hoặc x=2
a) 2x.(4x - 1)
câu b), c) mik ko biết
ko mong b cho mik
nhưng vẫn hi vọng b hoặc ai đó sẽ làm vậy
b) \(4x^4+1=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
a) \(8x^2-2x-1=8x^2-4x+2x-1=4x.\left(2x-1\right)+\left(2x-1\right)=\left(2x-1\right)\left(4x+1\right)\)
b) \(4x^4+1=\left(2x^2\right)^2+4x^2+1-4x^2=\left(2x^2+1\right)^2-4x^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
c) \(\left(x^2-2x\right)\left(x^2-2x-1\right)-6=x^4-2x^3-x^2-2x^3+4x^2+2x-6\)
\(=x^4-4x^3+3x^2+2x-6=\left(x^4-3x^3\right)-\left(x^3-3x^2\right)+\left(2x-6\right)\)
\(=x^3.\left(x-3\right)-x^2.\left(x-3\right)+2.\left(x-3\right)=\left(x-3\right).\left(x^3-x^2+2\right)\)
\(=\left(x-3\right)\left[\left(x^3+x^2\right)+\left(-2x^2-2x\right)+\left(2x+2\right)\right]\)
\(=\left(x-3\right)\left[x^2\left(x+1\right)-2x.\left(x+1\right)+2.\left(x+1\right)\right]=\left(x-3\right)\left(x+1\right)\left(x^2-2x+2\right)\)
a, 8x^2-2x-1 = 8x2-4x+2x-1 = 4x ( 2x -1) + (2x-1) = (4x+1)(2x-1)
b) 4x4+1 = (2x2)2 + 4x2+ 1 - 4x2 = (2x2+1)2-(2x)2 = (2x2+1-2x)(2x2+1+2x)
bài 1: a) \(x^2-3=x^2-\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\)
b) \(\left(a+b\right)^2-\left(a+b\right)^2=\left(a+b+a+b\right)\left(a+b-a-b\right)=2a+2b=2\left(a+b\right)\)
c) \(x^3-27b^3=\left(x-3b\right)\left(x^2+3xb+b^2\right)\)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
\(x^2+4x+4=\left(x+2\right)^2 \)
\(4x^2-4x+1=\left(2x-1\right)^2\)
\(c\left(x+1\right)-y\left(x+1\right)=\left(x+1\right)\left(c-y\right)\)
\(x^3-3x^2+3x-1+27y^3=\left(x-1\right)^3+27y^3=\left(x-1+3y\right)\left(x^2-2x+1-3xy+3y+9y^2\right)\)
1)x^4+x^2+1=x^4+2x^2+1-x^2
=(X^2+1)-x^2=(x^2+1-x)(x^2+1+1)
2)(b^2 + b + 1) ( b^2 + b + 2 ).
Biểu thức đã được biểu diễn dưới dạng các tích
3)x^4+4x^2-5
=x^4-x^2+5x^2-5
=(x^4-x^2)+(5x^2-5)
=x^2(x^2-1)+5(x^2-1)
=(x^2+5)(x^2-1)
=(x^2+5)(x+1)(x-1)