Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(16x^3-16x^4+4x-8x^2-1=0\)
<=> \(-16x^4-4x^2+16x^3+4x-4x^2-1=0\)
<=> \(-4x^2\left(4x+1\right)+4x\left(4x^2+1\right)-\left(4x^2+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(4x^2-4x+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(2x-1\right)^2=0\)
<=> \(2x-1=0\) (do 4x2 + 1 > 0 )
<=> \(x=\frac{1}{2}\)
a,X^3-16x =x(x^2-16)
b,y(y-2)-3(y-2)=(y+3).(y-2)
c,x^2+4x+4-y^2=(x+2)^2-y^2=(x+y+2).(x+2-Y)
D,4^2y^3-12x^2y^4+16X^5y^3=4x^2y^2(y-3y^2+4X^3y)
Dùng pp hệ số bất định nha bạn
Đặt
A = \(x^4-4x^3-7x^2+16x-3\)
A \(=\left(x^2+ax+1\right)\left(x^2+bx-3\right)\)
A \(=x^4+ax^3+x^2+bx^3+abx^2+bx-3x^2-3ax-3\)
A \(=x^4+\left(a+b\right)x^3+\left(1+ab-3\right)x^2+\left(b-3a\right)x-3\)
Đồng nhất 2 đa thức ta được :
\(\left\{{}\begin{matrix}a+b=-4\\-2+ab=-7\\b-3a=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=-5\\b=1\end{matrix}\right.\)
\(\Rightarrow A=\left(x^2-5x+1\right)\left(x^2+x-3\right)\)
\(=\left(4x^2-7x-50\right)^2-x^2\left(16x^2+56x+49\right)\)
\(=\left(4x^2-7x-50\right)^2-x^2\left(4x+7\right)^2\)
\(=\left(4x^2-7x-50-4x^2-7x\right)\left(4x^2-7x-50+4x^2+7x\right)\)
\(=\left(-14x-50\right)\left(8x^2-50\right)\)
\(=-4\left(7x+25\right)\left(2x-5\right)\left(2x+5\right)\)
Áp dụng hằng đẳng thức
a) x2+16x+64
=> x2+2.8x+82
=> (x+8)2
b) 25x2+10x+1
=> (5x+1)2
c) x2-12x+36
=> (x+6)2
d) 4x2-4x+1
=> (2x-1)2
e) x2-2x+1
=> (x-1)2
1.\(16x^3+54y^3=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^3\right)\)
2.\(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\)
3.\(16x-5x^2-3=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(x-3\right)\left(1-5x\right)\)
4.\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
5.\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
\(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2+y^2\right)\left(x^2-y^2\right)\)
\(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(\left(4x^4+4x^3-16x^2-16x\right):\left(4x+4\right)\)
\(=[\left(4x^4+4x^3\right)-\left(16x^2+16x\right)]:\left(4x+4\right)\)
\(=[4x^3.\left(x+1\right)-16x.\left(x+1\right)]:\left(4x+4\right)\)
\(=[\left(4x^3-16x\right).\left(x+1\right)]:\left(4x+4\right)\)
\(=[4.\left(x^3-4x\right).\left(x+1\right)]:[4.\left(x+1\right)]\)
\(=\frac{4.\left(x+1\right).\left(x^3-4x\right)}{4.\left(x+1\right)}\)
\(=x^3-4x\)