Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu này mih biết làm nhưng pp nhẩm nghiệm là sao bạn
bạn có thể cho mih vd đi\ược ko
a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
a)\(x^3-3x^2-4x+12\)
\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
b) \(x^4-5x^2+4\)
\(=\left(x^4-4x^2\right)-\left(x^2-4\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)\)
Bài làm
a) x3 - 12 + 3x2 - 4x
= ( x3 + 3x2 ) + ( -12 - 4x )
= x2( x + 3 ) - 4( x + 3 )
= ( x2 - 4 )( x + 3 )
b) 2x2 + 13x - 7
= 2x2 + 14x - 1x - 7
= ( 2x2 + 14x ) - ( x + 7 )
= 2x( x + 7 ) - ( x + 7 )
= ( 2x - 1 )( x + 7 )
# Học tốt #
a) \(12x^3+8x^2-3x-2=4x^2\left(3x+2\right)-\left(3x+2\right)\)
\(=\left(3x+2\right)\left(4x^2-1\right)=\left(3x+2\right)\left(2x-1\right)\left(2x+1\right)\)
b) \(18x^3+27x^2-2x-3=9x^2\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(9x^2-1\right)=\left(2x+3\right)\left(3x-1\right)\left(3x+1\right)\)
c) \(8x^3+4x^2-34x+15=4x^2\left(2x-3\right)+8x\left(2x-3\right)-5\left(2x-3\right)\)
\(=\left(2x-3\right)\left(4x^2+8x-5\right)=\left(2x-3\right)\left(2x-1\right)\left(2x+5\right)\)
1.a)\(x^2-ax+bx-ab=x\left(x-a\right)+b\left(x-a\right)=\left(x+b\right)\left(x-a\right)\)
b)\(x^2+ay-y^2-ax=\left(x-y\right)\left(x+y\right)-a\left(x-y\right)=\left(x+y-a\right)\left(x-y\right)\)
c)\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2.a)\(2x^2-12x=-18=>2x^2-12x+18=0=>x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)b)\(\left(4x^2-4x+1\right)-x^2=0=>3x^2-3x-x+1=3x\left(x-1\right)-\left(x-1\right)=\left(3x-1\right)\left(x-1\right)=0\)
\(=>\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}=>\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)
a) 2x2 - 12x = -18
<=> 2x2 - 12x + 18 = 0
<=> 2(x2 - 6x + 9) = 0
<=> 2(x2 - 2.x.3 + 9) = 0
<=> 2(x - 3)2 = 0
<=> x - 3 = 0
<=> x = 0 + 3
<=> x = 3
b) (4x2 - 4x + 1) - x2 = 0
<=> 4x2 - 4x + 1 - x2 = 0
<=> 3x2 - 4x + 1 = 0
<=> 3x2 - x - 3x + 1 = 0
<=> x(3x - 1) - (3x - 1) = 0
<=> \(\orbr{\begin{cases}\left(3x-1\right)=0\\\left(x-1\right)=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
a) \(2x^2-4xy+2y^2-8z^2=2\left(x^2-2xy+y^2-4z^2\right)=2\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=2\left(x-y-2z\right)\left(x-y+2z\right)\)
b) \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
a) \(a^3-b^3-3ab\left(a-b\right)\)
\(=a^3-3a^2b+3ab^2-b^3\)
\(=\left(a-b\right)^3\)
b) \(2x^3+x^2-4x-12\)
\(=2x^3-4x^2+5x^2-10x+6x-12\)
\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(2x^2+5x+6\right)\left(x-2\right)\)
c) \(x^3-3x^2+2\)
\(=x^3-x^2-2x^2+2x-2x+2\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x^2-2x-2\right)\left(x-1\right)\)
a) \(a^3-b^3-3ab\left(a-b\right)=\left(a-b\right)^3\)
b) \(2x^3+x^2-4x-12=2x^3-4x^2+5x^2-10x+6x-12\)
\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+5x+6\right)\)
c) \(x^3-3x^2+2=x^3-x^2-2x^2+2\)
\(=x^2\left(x-1\right)-2\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2+2x+2\right)\)