Question

mọi người giúp em với ạ

em cảm ơn mọi người rất nhiều ạ

Bài 1. Cho các đa thức: \( A(x) = x^4 - x^3 + x^2 - 3 \); \( B(x) = 3x^4 + x^3 - x^2 - 2 \); \( C(x) = x^4 + x^2 + \frac{3}{4} \)

Tính \( A(x) + B(x) \); \( B(x) - C(x) \); \( A(x) - B(x) + C(x) \) bằng hai cách.

Bài 2. Tìm đa thức \( P(x) \) biết:

a) \( P(x) + (x^3 - 2x^2 + 7x + 1) = x^3 - x^2 + 1 \);

b) \( P(x) - (2x^4 - x^3 + x - 3) = -x^4 - x^3 + 5x^2 + 4x + 2 \);

c) \( (5x^3 - 2x^2 + 4x + 4) - P(x) = 6x^3 - x^2 + 3x - 2 \)

Answer 1

Bài 1:

Cách 1: A(x)+B(x)

\(=x^4-x^3+x^2-3+3x^4+x^3-x^2-2=4x^4-5\)

Cách 2: A(x)+B(x)

\(=3x^4+x^3-x^2-2+x^4-x^3+x^2-3\)

\(=4x^4-5\)

Cách 1: B(x)-C(x)

\(=3x^4+x^3-x^2-2-x^4-x^2-\dfrac{3}{4}\)

\(=2x^4+x^3-2x^2-\dfrac{11}{4}\)

Cách 2: B(x)-C(x)

=-[C(x)-B(x)]

\(=-\left(x^4+x^2+\dfrac{3}{4}-3x^4-x^3+x^2+2\right)\)

\(=-\left(2x^4-x^3+2x^2+\dfrac{11}{4}\right)\)

\(=-2x^4+x^3-2x^2-\dfrac{11}{4}\)

Cách 1: A(x)-B(x)+C(x)

=A(x)-[B(x)-C(x)]

\(=\left(x^4-x^3+x^2-3\right)-\left(3x^4+x^3-x^2-2-x^4-x^2-\dfrac{3}{4}\right)\)

\(=x^4-x^3+x^2-3-2x^4-x^3+2x^2+\dfrac{11}{4}\)

\(=-x^4-2x^3+3x^2-\dfrac{1}{4}\)

Cách 2: A(x)-B(x)+C(x)

=[A(x)-B(x)]+C(x)

\(=\left(x^4-x^3+x^2-3-3x^4-x^3+x^2+2\right)+\left(x^4+x^2+\dfrac{3}{4}\right)\)

\(=\left(-2x^4-2x^3+2x^2-1\right)+x^4+x^2+\dfrac{3}{4}\)

\(=-x^4-2x^3+3x^2-\dfrac{1}{4}\)

Bài 2:

a: \(P\left(x\right)+\left(x^3-2x^2+7x+1\right)=x^3-x^2+1\)

=>\(P\left(x\right)=x^3-x^2+1-x^3+2x^2-7x-1\)

=>\(P\left(x\right)=x^2-7x\)

b: \(P\left(x\right)-\left(2x^4-x^3+x-3\right)=-x^4-x^3+5x^2+4x+2\)

=>\(P\left(x\right)=-x^4-x^3+5x^2+4x+2+2x^4-x^3+x-3\)

=>\(P\left(x\right)=x^4-2x^3+5x^2+5x-1\)

c: \(\left(5x^3-2x^2+4x+4\right)-P\left(x\right)=6x^3-x^2+3x-2\)

=>\(P\left(x\right)=5x^3-2x^2+4x+4-\left(6x^3-x^2+3x-2\right)\)

=>\(P\left(x\right)=5x^3-2x^2+4x+4-6x^3+x^2-3x+2\)

=>\(P\left(x\right)=-x^3-x^2+x+6\)