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\(P\left(x\right)-R\left(x\right)=6x^2-3x+2+3x^2-7x+5\)
\(=9x^2-10x+7\)
mà \(P\left(x\right)+R\left(x\right)=-x^2-4x+3\)
nên \(P\left(x\right)=\dfrac{9x^2-10x+7-x^2-4x+3}{2}\)
\(=\dfrac{8x^2-14x+10}{2}=4x^2-7x+5\)
\(R\left(x\right)=9x^2-10x+7-4x^2+7x-5=5x^2-3x+2\)
\(Q\left(x\right)=-3x^2+7x-5-5x^2+3x-2=-8x^2+10x-7\)
a,P (x)+Q (x)+Q (x)=(3x-2x2-2+6x3)+(3x2-x-2x3+4)+(1+4x3-2x)
=3x-2x2-2+6x3+3x2-x-2x3+4+1+4x3-2x
=(3x-x-2x)+(-2x2+3x2+3x2)+(-2+4+1)+(6x3-2x3+4x3)
=4x2+3+8x3
b,P (x)-Q (x)-R (x)=(3x-2x2-2+6x3)-(3x2-x-2x3+4)-(1+4x3-2x)
=3x-2x2-2+6x3-3x2+x+2x3-4-1+4x3-2x
=(3x +x-2x)+(-2x2-3x2)+(-2-4-1)+(2x3+4x3)
=2x-5x2-7 +6x3
\(P\left(x\right)+Q\left(x\right)-R\left(x\right)=2x^3+6x^2-5x+x^3-4x^3+3-5x^2+3x^3-x+4\)
\(=\left(2x^3+x^3-4x^3+3x^3\right)+\left(6x^2-5x^2\right)-\left(5x+x\right)+\left(3+4\right)\)
\(=2x^3+x^2-6x+7\)
Vậy \(P\left(x\right)+Q\left(x\right)-R\left(x\right)=2x^3+x^2-6x+7\)
a) P(x)=8x6-4x2+5x5-12x+7x2-2x5
=8x6+(-4x2+7x2)+(5x5-2x5)-12x
=8x6+3x2+3x5-12x
b) P(x)=8x6+3x2+3x5-12x
=8x6+3x5+3x2-12x
P(x)-Q(x)=(8x6+3x5+3x2-12x)-(2x5-6x2+8x-2x6)
=8x6+3x5+3x2-12x-2x5+6x2-8x+2x6
=(8x6+2x6)+(3x5-2x5)+(3x2+6x2)+(-12x-8x)
=10x6+x5+9x2-20x
R(x)-Q(x)=4x6-8x2
R(x) =(4x6-8x2)+Q(x)
R(x) =(4x6-8x2)+(2x5-6x2+8x-2x6)
R(x) =4x6-8x2+2x5-6x2+8x-2x6
R(x) =(4x6-2x6)+(-8x2-6x2)+2x5+8x
R(x) =2x6-14x2+2x5+8x
\(P\left(x\right)=4x^4+2x^2-8x+\dfrac{1}{2}\)
\(Q\left(x\right)=-x^4-5x^2-8x-\dfrac{3}{4}\)
a: \(R\left(x\right)=P\left(x\right)-Q\left(x\right)=3x^4+7x^2+\dfrac{5}{4}\)
b: \(R\left(x\right)=3x^4+7x^2+\dfrac{5}{4}\ge\dfrac{5}{4}\forall x\)
nên R(X) không có nghiệm
Cộng 3 đẳng thức vế với vế ta có:
\(2\left(P\left(x\right)+Q\left(x\right)+R\left(x\right)\right)=6x^2-3x+2-3x^2+7x-5-x^2-4x+3\)
=>\(2\left(P\left(x\right)+Q\left(x\right)+R\left(x\right)\right)=2x^2\)
=> \(P\left(x\right)+Q\left(x\right)+R\left(x\right)=x^2\)
=>\(\hept{\begin{cases}R\left(x\right)=x^2-\left(6x^2-3x+2\right)=-5x^2+3x-2\\P\left(x\right)=x^2-\left(-3x^2+7x-5\right)=4x^2-7x+5\\Q\left(x\right)=x^2-\left(-x^2-4x+3\right)=2x^2+4x-3\end{cases}}\)