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a) Ta có : \(A\left(x\right)+B\left(x\right)\)
\(=2x^3+2x-3x^2+1+2x^2+3x^3-x-5\)
\(=\left(2x^3+3x^3\right)+\left(-3x^2+2x^2\right)+\left(2x-x\right)+\left(1-5\right)\)
\(=5x^3-x^2-x-4\)
b) Ta sẽ sắp xếp như sau :
\(A\left(x\right)=2x^3-3x^2+2x+1\)
\(B\left(x\right)=3x^3+2x^2-x-5\)
c) Ta có : \(A\left(x\right)-B\left(x\right)\)
\(=\left(2x^3+2x-3x^2+1\right)-\left(2x^2+3x^3-x-5\right)\)
\(=2x^3+2x-3x^2+1-2x^2-3x^3+x+5\)
\(=\left(2x^3-3x^3\right)+\left(-3x^2-2x^2\right)+\left(2x+x\right)+\left(1+5\right)\)
\(=-x^3-5x^2+3x+6\)
a) A(x) = 2x–3x2–3+4x3–x2–2x–5 = \(4x^3-4x^2-4x-8.\)
B(x) = 3x–4x3–1+3x2–5x–3x2\(=-4x^3-2x-1\)
b) M(x) = A(x) + B(x) \(=-4x^2-6x-9\)
c) Để M(x) = –9 => M(x) = \(=-4x^2-6x-9\)= -9
\(=-4x^2-6x=0\)
\(\Leftrightarrow-2x\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-2x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\2x=3\Leftrightarrow x=\frac{3}{2}\end{cases}}}\)
d) Ta có: đa thức K(x) = 5x–1
\(\Leftrightarrow K\left(x\right)=5x-1=0\)
\(\Leftrightarrow5x=1\)
\(\Leftrightarrow x=\frac{1}{5}\)
Vậy....
a)\(A\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6\\ B\left(x\right)=x^5+2x^4-2x^3+3x^2-x+\frac{1}{4}\)
b)\(A\left(x\right)+B\left(x\right)\)
\(\left(5x^5-4x^4-2x^3+4x^2+3x+6\right)+\left(x^5+2x^4-2x^3+3x^2-x+\frac{1}{4}\right)\\ =5x^2-4x^4-2x^3+4x^2+3x+6+x^5+2x^4-2x^3+3x^2-x+\frac{1}{4}\\ =\left(5x^5+x^5\right)+\left(-4x^4+2x^4\right)+\left(-2x^3-2x^3\right)+\left(4x^2+3x^2\right)+\left(3x-x\right)+\left(6+\frac{1}{4}\right)\\ =6x^5-2x^4-4x^3+7x^2+2x+\frac{25}{4}\)
trắc nghiệm
câu 1: c
câu 2: B
câu 3: D
câu 4: A
câu 5: C
câu 6: D
tự luận
câu 1:
a)M(x) = x4 + 2x2 + 1
b) M(x) + N(x) = -4x4 + x3 + 5x2 - 2
M(x) - N(x) = 6x4 - x3 - x2 + 4
c) \(M\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^4+2\left(-\dfrac{1}{2}\right)^2+1=\dfrac{25}{16}\)
I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
a, Sắp xếp : \(P\left(x\right)=2x^3+5x^2-3x^4+7-4x\)
\(\Rightarrow P\left(x\right)=-3x^4+2x^3-5x^2-4x+7\)
\(Q\left(x\right)=-3+2x^4-x+x^3-5x^2\)
\(\Rightarrow Q\left(x\right)=2x^4+x^3-5x^2-x-3\)
b, Ta có :* Đặt \(V\left(x\right)=P\left(x\right)+Q\left(x\right)\)
hay \(V\left(x\right)=2x^3+5x^2-3x^4+7-4x-3+2x^4-x+x^3-5x^2\)
\(=3x^3-x^4+4-5x\)
Vậy \(V\left(x\right)=3x^3-x^4+4-5x\)
Ta có : * Đặt \(K\left(x\right)=P\left(x\right)-Q\left(x\right)\)
hay \(2x^3+5x^2-3x^4+7-4x-\left(-3+2x^4-x+x^3-5x^2\right)\)
\(=2x^3+5x^2-3x^4+7-4x+3-2x^4+x-x^3+5x^2\)
\(=x^3+10x^2-5x^4+10-3x\)
Vậy \(K\left(x\right)=x^3+10x^2-5x^4+10-3x\)
Trả lời:
a, P(x) = - 3x2 + 3x - ( - 4x3 ) + 5 - (- 2x4 ) - x + 1
= - 3x2 + 3x + 4x3 + 5 + 2x4 - x + 1
= 2x4 + 4x3 - 3x2 + 2x + 6
Q(x) = 5x4 + 19x2 + 4x3 - ( - 6x ) - 12 - x2 - ( - 1 )
= 5x4 + 19x2 + 4x3 + 6x - 12 - x2 + 1
= 5x4 + 4x3 + 18x2 + 6x + 1
b, P(x) + Q(x) = 2x4 + 4x3 - 3x2 + 2x + 6 + 5x4 + 4x3 + 18x2 + 6x + 1
= 7x4 + 8x3 + 15x2 + 8x + 7
c, P(x) - Q(x) = 2x4 + 4x3 - 3x2 + 2x + 6 - ( 5x4 + 4x3 + 18x2 + 6x + 1 )
= 2x4 + 4x3 - 3x2 + 2x + 6 - 5x4 - 4x3 - 18x2 - 6x - 1
= - 3x4 - 21x3 - 4x + 5
a; A(\(x\)) = \(x^5\) - 2\(x^4\) + \(x^2\) - \(x\) + 1
A(\(x\)) = \(x^5\) - 2\(x^4\) + \(x^2\) - \(x\) + 1
B(\(x\)) = 6 - 2\(x\) - 3\(x^3\) + \(x^4\) - 3\(x^5\)
B(\(x\)) = -3\(x^5\) + \(x^4\) - 3\(x^3\) - 2\(x\) + 6
b; A(\(x\)) + B(\(x\)) = \(x^5\) - 2\(x^4\) + \(x^2\) - \(x\) + 1 + \(x^4\) - 3\(x^5\) - 3\(x^3\) - 2\(x\) + 6
A(\(x\)) + B(\(x\)) = (\(x^5\) - 3\(x^5\)) - (2\(x^4\) - \(x^4\)) - 3\(x^3\) + \(x^2\) - (\(x+2x\)) + (1+6)
A(\(x\)) + B(\(x\)) = -2\(x^5\) - \(x^4\) - 3\(x^3\) + \(x^2\) - 3\(x\) + 7