6x3=
Phép nhân 3x(2x2 – 4x + 1) được kết quả là:
A.6x3 – 4x + 1
B.6x3 – 12x2 + 3x
C.6x3 + 12x2 – 3x
D.6x3 + 12x2 + 3x
B.6x3-12x2+3x
chac zay
Phân tích đa thức thành nhân tử:
a) x 4 - 6 x 3 + 12 x 2 - 14x + 3.
b) x 4 + 6 x 3 + 7 x 2 -6x + l.
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
a.11/12+5/6x3/2-2/3
b.4/9:4/6x3/4+5/3
c.5/7:11/14:15/22x3/5
a) 11/12 + 5/6 x 3/2 - 2/3
= 11/12 + 15/12 - 2/3
= 26/12 - 2/3
= 26/12 - 8/12
= 18 /12 = 3/2
b) 4/9 : 4/6 x 3/4 +5/3
= 4/9 x 6/4 x 3/4 + 5/3
= 2/3 x 3/4 + 5/3
= 6/12 + 5/3
= 6/12 + 20/12
= 26/12 = 13/6
c) 5/7 : 11/14 : 15/22 x 3/5
= 5/7 x 14/11 x 22/15 x 3/5
= 10/11 x 22/15 x 3/5
= 4/3 x 3/5 = 20/15 = 4/3
a. \(\dfrac{11}{12}+\dfrac{5}{6}\)x\(\dfrac{3}{2}-\dfrac{2}{3}\) \(=\)\(\dfrac{11}{12}+\dfrac{5}{4}-\dfrac{2}{3}=\dfrac{11}{12}+\dfrac{15}{12}-\dfrac{8}{12}=\dfrac{11+15-8}{12}=\dfrac{18}{12}=\dfrac{3}{2}\)
b. \(\dfrac{4}{9}:\dfrac{4}{6}\)x\(\dfrac{3}{4}+\dfrac{5}{3}\) \(=\dfrac{4}{9}:\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{9}+\dfrac{5}{3}=\dfrac{8}{9}+\dfrac{15}{9}=\dfrac{8+15}{9}=\dfrac{23}{9}\)
c. \(\dfrac{5}{7}:\dfrac{11}{14}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{10}{14}:\) \(\dfrac{11}{14}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{10}{11}\)\(:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{20}{22}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{4}{3}\)x\(\dfrac{3}{5}=\dfrac{4}{5}\)
P(x)=-6x3-2+4x2+2x-2
Q(x)=-8-4x2+6x3-x4+3x
a,xắp sếp từ lớn đến nhỏ theo lũy thừa
b,tính P(x)-Q(x);P(x)+Q(x)
a: \(P\left(x\right)=6x^3+4x^2+2x-4\)
\(Q\left(x\right)=-x^4+6x^3-4x^2+3x-8\)
b: \(P\left(x\right)-Q\left(x\right)=x^4+8x^2-x+4\)
\(P\left(x\right)+Q\left(x\right)=-x^4+12x^3+5x-12\)
Bài 3:(2,5đ) Cho các đa thức
M(x) = 6x3 – 2x2 + 3x +10
N(x) = –6x3 + x2 – 6x -10
a) Tính M(x) + N(x); M(x) – N(x)
b) Tìm nghiệm của đa thức M(x)+N(x)
a. M(x) + N(x) = 6x3 – 2x2 + 3x +10 - 6x3 + x2 – 6x -10
= (6x3 - 6x3 ) + ( -2x2 + x2 ) + ( 3x - 6x ) + ( 10 - 10 )
= -x2 - 3x
M(x) - N(x) = 6x3 – 2x2 + 3x +10 - ( –6x3 + x2 – 6x -10)
= 6x3 – 2x2 + 3x +10 + 6x3 - x2 + 6x +10
= (6x3 + 6x3 ) + ( -2x2 - x2 ) + ( 3x + 6x) + ( 10 + 10)
= 12x3 - 3x2 + 9x + 20
b. Đặt -x2 - 3x = 0
=> -x2 + (-3)x = 0
=> -x2 + 3.-x = 0
=> -x(-x+ 3) = 0
=>\(\left[{}\begin{matrix}-x=0\\-x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\-x=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy nghiệm của đa thức trên là 0 hoặc -3
a) M(X) + N(x)= (6x3 – 2x2 + 3x +10)
+ (–6x3 + x2 – 6x -10)
M(x) + N(x)= – x2 - 3x.
M(x) + N(x)= (6x3 – 2x2 + 3x +10)
- (–6x3 + x2 – 6x -10)
M(x) - N(x)= 12x3 - x2 + 9x + 20.
b) Nghiệm của M(x) + N(x)= x= 0, -3.
What's 6x3 ?
5/6x3/5
\(\frac{5}{6}x\frac{3}{5}=\frac{1x3}{6x1}=\frac{3}{6}=\frac{1}{2}\)
duyệt đi
−6x3+4x2+8x4−3x+5
(6x3 - 7x2 - x + 2) : (2x + 1)
\(=\left(6x^3+3x^2-10x^2-5x+4x+2\right):\left(2x+1\right)\\ =\left[3x^2\left(2x+1\right)-5x\left(2x+1\right)+2\left(2x+1\right)\right]:\left(2x+1\right)\\ =3x^2-5x+2\)
\(=3x^2-\dfrac{7}{2}x-2+x+6x^3-7x^2+x+2\\ =-4x^2-\dfrac{7}{2}x+6x^3\)