X4 + 4 = 5X( X2 - 2)
Tìm đa thức M biết:
a) 2 x 6 - x 4 - 2 x 2 +1 = M.(2 x 2 -1);
b) ( x 2 +x + 1).M = x 4 - x 3 - 4 x 2 - 5x - 3.
a) Kết quả M = x 4 – 1.
b) Kết quả M = x 2 – 2x – 3.
cho phương trình \(x^4-5x^2+3=0\) có 4 nghiệm là x1 ; x2 ; x3; x4
Khi đó x1 +x2 +x3 +x4 =? ...................Ho mk cai
đặt x^2=t =>pt<=> t^2-5t+3=0 =>đen ta =25-12=13
giai ra t rồi tìm x có 4nghiệm
Phân tích đa thức thành nhân tử:
a) x 4 + 1 - 2 x 2 ; b) x 2 - y 2 - 5y + 5x;
c) y 2 - 4 x 2 +4x - 1; d) x3 ( 2 + x ) 2 - ( x + 2 ) 2 + 1 - x 3 .
Phân tích đa thức thành nhân tử:
a) x 2 -3x + 2; b) 4 x 2 - 36x + 56;
c) 2 x 2 + 5x + 2; d)2 x 2 -9x + 7;
e) 4 x 2 - 4x - 9 y 2 + 12y - 3; g) x 4 - 2 x 3 -4 x 2 + 4x-3;
h) x 3 -x +3 x 2 y + 3x y 2 + y 3 -y.
a) (x - 1)(x - 2). b) 4(x - 2)(x - 7).
c) (x + 2)(2x +1). d) (x - l)(2x - 7).
e) (2x + 3y - 3)(2x - 3y +1). g) (x - 3)( x 3 + x 2 - x +1).
h) (x + y)(x + y-l)(x + y + l).
Cho hai đa thức
A ( x ) = x 5 + x 2 + 5 x + 6 - x 5 - 3 x - 5 , B ( x ) = x 4 + 2 x 2 - 3 x - 3 - x 4 - x 2 + 3 x + 4
b. Tính A ( x ) + B ( x ) v à A ( x ) - B ( x )
b. Ta có:
A(x) + B(x) = x2 + 2x + 1 + x2 + 1 = 2x2 + 2x + 2 (0.5 điểm)
A(x) - B(x) = x2 + 2x + 1 - (x2 + 1) = 2x (0.5 điểm)
Phân tích đa thức thành nhân tử:
a) xy + y2 – x – y
b) 25 – x2 + 4xy – 4y2
c) 4x3 + 4xy2 + 8x2y – 16x
d) (x2 + x)2 + 4(x2 + x) – 12
e) (x + 1) (x + 2) (x + 3) (x + 4) - 24 g)
h) x2 – 5x + 4
i) x4 – 5x2 + 4
j) x3 – 2x2 + 6x – 5
k) x2 – 4x + 3
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
phân tích đa thức thành nhân tử
a) x2- x- y2- y
b) x2- 2xy- y2-z2
c) 5x- 5y+ 4x- ay
d) 3x3- x2-21x+ 7
e) x3- 4x2- 8x- 8
f) x3- 5x2- 5x+ 1
g) x2y- xz+ z- y
h) x4- x3+ x2- 1
i) x4- x2+ 10x- 25
a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
Thực hiện phép tính:
a)(x4-3x-1):(x2-x-1)
b)(x3-x2+5x-4):(-x+2x2+1)
c)(2x2+2x-5x3+2x4-1):(-x+x2+1)
\(a,=\left[x^2\left(x^2-x-1\right)+x^3+x^2-3x-1\right]:\left(x^2-x-1\right)\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2x^2-2x-1\right]\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)\\ =\left[\left(x^2+x+2\right)\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)=x^2+x+2R1\)
7) x4+2x3-2x2+2x-3=0
8) (x-1)( x2+5x-2)-x3+1=0
9) x2+(x+2)(11x-7)=4
(GIẢI PHƯƠNG TRÌNH)
\(x^4+2x^3-2x^2+2x-3=0\\ \Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\\ \Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^3-x^2+x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\left(\text{vì }x^2+1\ge1>0\right)\)
Vậy ...
\(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^2+5x-2\right)-\left(x^2+x+1\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
\(x^2+\left(x+2\right)\left(11x-7\right)=4\\ \Leftrightarrow x^2-4+\left(x+2\right)\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2\right)+\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2+11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...