cho a=x2 + x +1
tinh theo a gt cau x4 +2x3+5x2+4x+4
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b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=5x^3+14x^2+12x+8\)
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
a) Ta có: B(x)-M(x)=A(x)
nên M(x)=B(x)-A(x)
\(=x^4-2x^3+5x^2+x+10-x^4-2x^3+5x^2+3x+6\)
\(=-4x^3+10x^2+4x+16\)
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Thu gọn và sắp xếp phải k ạ?
`F(x)= (x^4-x^4)+(5x^2-8x^2)-4x+x^5+3+2x^3+2`
`F(x) = -3x^2-4x+x^5+3+2x^3+2`
`F(x)= x^5+2x^3-3x^2-4x+3+2`
\(F\left(x\right)=x^4+5x^2-4x+x^5-x^4-8x^2+3+2x^3+2\)
\(F\left(x\right)=x^5+\left(x^4-x^4\right)+2x^3+\left(5x^2-8x^2\right)-4x+\left(3+2\right)\)
\(F\left(x\right)=x^5+2x^3-3x^2-4x+5\)
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)
\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)
\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)
\(=5x^2+4x+4\)
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
Trước hết, ta rút gọn các đa thức:
- Q(x) = 4x3 – 2x + 5x2 - 2x3 + 1 - 2x3
Q(x) = (4x3- 2x3- 2x3) – 2x + 5x2 + 1
Q(x) = 0 – 2x + 5x2 + 1
Q(x) = – 2x + 5x2 + 1
- R(x) = - x2 + 2x4 + 2x - 3x4 – 10 + x4
R(x) = - x2 + (2x4- 3x4+ x4) + 2x – 10
R(x) = - x2 + 0 + 2x – 10
R(x) = - x2 + 2x – 10
Sắp xếp các hạng tử của đa thức sau theo lũy thừa giảm dần của biến ta có:
Q(x) = 5x2 – 2x + 1
R(x) = - x2 + 2x – 10
a: \(A\left(x\right)=9-x^5+4x-2x^3+x^2-7x^4\)
\(=-x^5-7x^4-2x^3+x^2+4x+9\)
\(B\left(x\right)=x^5-9+2x^2+7x^4+2x^3-3x\)
\(=x^5+7x^4+2x^3+2x^2-3x-9\)
b: A(x)+B(x)
\(=-x^5-7x^4-2x^3+x^2+4x+9+x^5+7x^4+2x^3+2x^2-3x-9\)
\(=3x^2+x\)
A(x)-B(x)
\(=-x^5-7x^4-2x^3+x^2+4x+9-x^5-7x^4-2x^3-2x^2+3x+9\)
\(=-2x^5-14x^4-4x^3-x^2+7x+18\)
x4 +2x3+5x2+4x+4 = (x2 + x + 2)2 = (a + 1)2
sai roi ban