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A(x) = 3x^3 - 4x^4 - 2x^3 + 4x^4 - 5x + 3
= (4x^4 - 4x^4) + (3x^3 - 2x^3) - 5x + 3
= x^3 - 5x + 3
B(x) = 5x^3 - 4x^2 - 5x^3 - 4x^2 - 5x - 3
= (5x^3 - 5x^3) - (4x^2 + 4x^2) - 5x - 3
= -8x^2 - 5x - 3
b.
\(A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\)
\(=x^3-8x^2+\left(-5x-5x\right)+\left(3-3\right)\)
\(=x^3-8x^2-10x\)
\(A\left(x\right)-B\left(x\right)=\left(x^3-5x+3\right)-\left(-8x^2-5x-3\right)\)
\(=x^3-5x+3+8x^2+5x+3\)
\(=x^3-8x^2+\left(-5x+5x\right)+\left(3+3\right)\)
\(=x^3-8x^2+6\)
Vậy \(A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\)
\(A\left(x\right)-B\left(x\right)=x^3-8x^2+6\)
a) \(A\left(x\right)=3x^3-4x^4-2x^3+4x^4-5x+3\)
\(\Rightarrow A\left(x\right)=-4x^4+4x^4+3x^3-2x^3-5x+3\)
\(\Rightarrow A\left(x\right)=x^3-5x+3\)
\(B\left(x\right)=5x^3-4x^2-5x^3-4x^2-5x-3\)
\(\Rightarrow B\left(x\right)=5x^3-5x^3-4x^2-4x^2-5x-3\)
\(\Rightarrow B\left(x\right)=-8x^2-5x-3\)
b) \(A\left(x\right)+B\left(x\right)=x^3-5x+3+\left(-8x^2-5x-3\right)\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-5x-5x+3-3\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\)
\(A\left(x\right)-B\left(x\right)=x^3-5x+3-\left(-8x^2-5x-3\right)\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3-5x+3+8x^2+5x+3\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2-5x+5x+3+3\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2+6\)
\(M\left(x\right)=3x^4-2x^3+5x^2-4x+1\)
\(N\left(x\right)=-3x^4+2x^3-5x^2+7x+5\)
\(P\left(x\right)=M\left(x\right)+N\left(x\right)\)
\(=\left(3x^4-2x^3+5x^2-4x+1\right)+\left(-3x^4+2x^3-5x^2+7x+5\right)\)
\(=3x+6\)
\(Q\left(x\right)=M\left(x\right)-N\left(x\right)\)
\(=\left(3x^4-2x^3+5x^2-4x+1\right)-\left(-3x^4+2x^3-5x^2+7x+5\right)\)
\(=3x^4-2x^3+5x^2-4x+1+3x^4-2x^3+5x^2-7x-5\)
\(=6x^4-4x^3+10x^2-11x-4\)
a: \(P\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6\)
Bậc là 5
\(Q\left(x\right)=-5x^5+4x^4+2x^3-4x^2+7x+\dfrac{1}{4}\)
Bậc là 5
b: H(x)=P(x)+Q(x)
\(=5x^5-4x^4-2x^3+4x^2+3x+6-5x^5+4x^4+2x^3-4x^2+7x+\dfrac{1}{4}\)
=10x+6,25
c: Để H(x)=0 thì 10x+6,25=0
hay x=-0,625
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)\(A\left(x\right)+B\left(x\right)\)
\(\left(5x^5-4x^4-2x^3+4x^2+3x+6\right)+\left(x^5+2x^4-2x^3+3x^2-x+\frac{1}{4}\right)\\ =5x^2-4x^4-2x^3+4x^2+3x+6+x^5+2x^4-2x^3+3x^2-x+\frac{1}{4}\\ =\left(5x^5+x^5\right)+\left(-4x^4+2x^4\right)+\left(-2x^3-2x^3\right)+\left(4x^2+3x^2\right)+\left(3x-x\right)+\left(6+\frac{1}{4}\right)\\ =6x^5-2x^4-4x^3+7x^2+2x+\frac{25}{4}\)
a) Ta có: \(f\left(x\right)=5x-3x^2+2x^4-3x-x^4-5\)
\(=x^4-3x^2+2x-5\)
Ta có: \(g\left(x\right)=2x^3+10x-1-7x^2-15x+10x^2\)
\(=2x^3+3x^2-5x-1\)
b) Ta có: f(x)+g(x)
\(=x^4-3x^2+2x-5+2x^3+3x^2-5x-1\)
\(=x^4-2x^3-3x-6\)
Ta có: f(x)-g(x)
\(=x^4-3x^2+2x-5-2x^3-3x^2+5x+1\)
\(=x^4-2x^3-6x^2+7x-4\)
bài 3:
a) f(x)= x2+2x4-2x3+x2+5x4+4x3-x+5
= (2x4+5x4)+(4x3-2x3)+(x2+x2)-x+5
= 7x4+2x3+2x2-x+5
g(x)= -2x2+8x4+x-x4-3x3+3x2+5+4x3
=(8x4-x4)+(4x3-3x3)+(3x2-2x2)+x+5
= 7x4+x3+x2+x+5
b) h(x)=f(x)-g(x)
=(7x4+2x3+2x2-x+5)-(7x4+x3+x2+x+5)
=7x4+2x3+2x2-x+5-7x4-x3-x2-x-5
=(7x4-7x4)+(2x3-x3)+(2x2-x2)-(x+x)+(5-5)
=x3+x2-2x
Bài 4:
a) f(x)=5x4+x3-x+11+x4-5x3
=(5x4+x4)+(x3-5x3)-x+11
=6x4-4x3-x+11
g(x)=2x3+3x4+9-4x3+2x4-x
=(3x4+2x4)+(2x3-4x3)-x+9
=5x4-2x3-x+9
b) h(x)=f(x)-g(x)
=(6x4-4x3-x+11)-(5x4-2x3-x+9)
=6x4-4x3-x+11-5x4-2x3-x+9
=(6x4-5x4)-(4x3+2x3)-(x+x)+(11+9)
= x4-6x3-2x+20
c) Với x = -2
Ta có: h(-2)=(-2)4-6.(-2)3-2.(-2)+20=88\(\ne\)0
Vậy x = -2 không phải là nghiệm của đa thức h(x)
đúng thì tặng 1 tick cho mk nk các pn!!!
a: \(P\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}\)
\(Q\left(x\right)=4x^4+2x^3-5x^2-6x+\dfrac{3}{2}\)
b: \(A\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}+4x^4+2x^3-5x^2-6x+\dfrac{3}{2}=-x^4+2x^3-3x^2-14x+2\)
\(B\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}-4x^4-2x^3+5x^2+6x-\dfrac{3}{2}=-9x^4-2x^3+7x^2-2x-1\)
a)\(A\left(x\right)=x^4+4x^3+2x^2+x-7\)
\(B\left(x\right)=2x^4-4x^3-2x^2-5x+3\)
b) \(f\left(x\right)=A\left(x\right)+B\left(x\right)=x^4+4x^3+2x^2+x-7+2x^4-4x^3-2x^2-5x+3=3x^4-4x-4\)
\(g\left(x\right)=A\left(x\right)-B\left(x\right)=x^4+4x^3+2x^2+x-7-2x^4+4x^3+2x^2+5x-3=-x^4+8x^3+4x^2+6x-10\)c)\(g\left(0\right)=-0^4+8.0^3+4.0^2+6.0-10=-10\)
\(g\left(-2\right)=\left(-2\right)^4+8.\left(-2\right)^3+4.\left(-2\right)^2+6.\left(-2\right)-10=16-64+16-12-10=-54\)
a) \(A\left(x\right)=-2x+3x-4x^3+\frac{3}{5}-5x^4\)
\(\Rightarrow A\left(x\right)=-5x^4-4x^3-2x+3x+\frac{3}{5}\)
\(\Rightarrow A\left(x\right)=-5x^4-4x^3+x+\frac{3}{5}\)
\(B\left(x\right)=3x^4+\frac{1}{5}-7x^2+5x^3-9x\)
\(\Rightarrow B\left(x\right)=3x^4+5x^3-7x^2-9x+\frac{1}{5}\)
b) \(A\left(x\right)+B\left(x\right)\)
\(=-5x^4-4x^3+x+\frac{3}{5}+3x^4+5x^3-7x^2-9x+\frac{1}{5}\)
\(=\left(-5x^4+3x^4\right)+\left(-4x^3+5x^3\right)-7x^2+\left(x-9x\right)+\left(\frac{3}{5}+\frac{1}{5}\right)\)
\(=-2x^4+x^3-7x^2-8x+\frac{4}{5}\)
\(A\left(x\right)-B\left(x\right)\)
\(=\left(-5x^4-4x^3+x+\frac{3}{5}\right)-\left(3x^4+5x^3-7x^2-9x+\frac{1}{5}\right)\)
\(=\left(-5x^4-3x^4\right)+\left(-4x^3-5x^3\right)+7x^2+\left(x+9x\right)+\left(\frac{3}{5}-\frac{1}{5}\right)\)
\(=-8x^4-9x^3+7x^2+10x+\frac{2}{5}\)
Chúc bạn học tốt !