giải pt : x4+2x3+5x2+4x-12=0
Giải Phương trình
5x2 + 4x + 2x3 + x4 - 12 = 0
\(5x^2+4x+2x^3+x^4-12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^3+2x^2+x^2+2x+6x+12\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[x^2+2\times\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+6\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vì \(\left(x^2+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\)
\(\Rightarrow\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) vô nghiệm
Vậy phương trình có tập nghiệm là\(S=\left\{1;-2\right\}\)
F(x)=x4+5x2-4x+x5-x4-8x2+3+2x3+2
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`
\(F\left(x\right)=x^4+5x^2-4x+x^5-x^4-8x^2+3+2x^3+2\)
\(F\left(x\right)=x^5+\left(x^4-x^4\right)+2x^3+\left(5x^2-8x^2\right)-4x+\left(3+2\right)\)
\(F\left(x\right)=x^5+2x^3-3x^2-4x+5\)
Phân tích
a,(x2 + x + 2)3 - (x+1)3 = x6 +1 b,(x2 + 10x + 8)2 - (8x + 4)(x2 + 8x+7)
c, A= x4 + 2x3 + 3x2 + 2x+4 d,B= x4 + 4x3 + +8x2 + 8x + 4
e, C= x4 - 2x3 + 5x2 - 4x + 4
Giải pt
X4+2x3-6x2-2x+1=0
A(x)=x4+2x3-5x2-3x-6
B(x)=-x4-2x3+5x2+x+10
a/Tìm đa thức M(x) sao cho B(x)-M(x)=A(x)
a) Ta có: B(x)-M(x)=A(x)
nên M(x)=B(x)-A(x)
\(=x^4-2x^3+5x^2+x+10-x^4-2x^3+5x^2+3x+6\)
\(=-4x^3+10x^2+4x+16\)
Giải phương trình:
a) x4 - 2x3 + x2 - 4x +4 = 0
b) x4 + 2x3 - 3 = 0
c) 2x4 - 100x + 98 = 0
d) (x + 1)(x + 2)(x + 3)(x + 4) = 24
d: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Bài toán 3 : Thực hiện phép chia.
a) (2x3 – 5x2 – x + 1) : (2x + 1)
b) (x3 – 2x + 4) : (x + 2)
c) (6x3 – 19x2 + 23x – 12) : (2x – 3)
d) (x4 – 2x3 – 1 + 2x) : (x2 – 1)
a: \(\dfrac{2x^3-5x^2-x+1}{2x+1}\)
\(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}\)
\(=x^2-3x+1\)
b: \(\dfrac{x^3-2x+4}{x+2}\)
\(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}\)
\(=x^2-2x+2\)
thực hiện phép chia
(-3x3 + 5x2 - 9x + 15) : (-3 + 5)
(x4 - 2x3 + 2x -1) : (x2 - 1)
(5x4 + 9x3 - 2x2 - 4x -8) : (x-1)
(5x3 + 14x2 + 12x + 8) : (x+2)
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: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)
\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)
\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)
\(=5x^2+4x+4\)
thực hiện phép chia
(-3x3 + 5x2 - 9x + 15) : (-3 + 5)
(x4 - 2x3 + 2x -1) : (x2 - 1)
(5x4 + 9x3 - 2x2 - 4x -8) : (x-1)
(5x3 + 14x2 + 12x + 8) : (x+2)
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
a)(-3x2+5x2-9x+15):(-3x+5)
b)(x4-2x3+2x-1):(x2-1)
c)(5x4+9x3-2x2-4x-8):(x-1)
d)(5x3+14x2+12x+8):(x+2)
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: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=5x^3+14x^2+12x+8\)