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a) \(A=\)\(x^4\)\(+4x^3\)\(+2x^2\)\(+x\)\(-7\)
\(B=\)\(2x^4\)\(-4x^3\)\(-2x^2\)\(-5x\)\(+3\)
b) f(x)= A(x)+B(x)= \(3x^4-4x\)\(-4\)
g(x)=A(x)-B(x) = \(-x^4+8x^3+4x^2+6x\)\(-10\)
c) g(x)= \(0^4+8.0^3+4.0^2\)\(+6.0\)\(-10\)
= -10
g(-2)=\(-2^4+8.-2^3+4.-2^2+6.-2\)\(-10\)
=\(-54\)
Lời giải:
a)
$M(x)=(x^5+5x^5)-2x^4-4x^3+3x$
$=6x^5-2x^4-4x^3+3x$
$N(x)=-6x^5+(7x^4-5x^4)+(x^3+3x^3)+4x^2-3x-1$
$=-6x^5+2x^4+4x^3+4x^2-3x-1$
b)
$M(-1)=6(-1)^5-2(-1)^4-4(-1)^3+3(-1)=-7$
$N(-2)=-6(-2)^5+2(-2)^4+4(-2)^3+4(-2)^2-3(-2)-1$
$=213$
c)
$M(x)+N(x)=(6x^5-2x^4-4x^3+3x)+(-6x^5+2x^4+4x^3+4x^2-3x-1)$
$=4x^2-1$
$M(x)-N(x)=(6x^5-2x^4-4x^3+3x)-(-6x^5+2x^4+4x^3+4x^2-3x-1)$
$=12x^5-4x^4-8x^3-4x^2+6x+1$
d)
$F(x)=M(x)+N(x)=4x^2-1=0\Leftrightarrow x^2=\frac{1}{4}$
$\Leftrightarrow x=\pm \frac{1}{2}$
Vậy $x=\pm \frac{1}{2}$ là nghiệm của $F(x)$
Ta có :
A(x) = 3x - 2x2 - 2 +6x2 = 4x2 + 3x - 2
B(x) = 3x2 - x - 2x3 + 4 = -2x3 + 3x2 - x + 4
C(x) = 1 + 4x3 - 2x = 4x3 - 2x + 1
⇒ A(x) + B(x) - C(x)
= (4x2 + 3x - 2) + (-2x3 + 3x2 - x + 4) - (4x3 - 2x + 1)
= 4x2 + 3x - 2 - 2x3 + 3x2 - x + 4 - 4x3 + 2x - 1
= 7x2 + 4x + 1 - 6x3 = -6x3 + 7x2 + 4x + 1
bài 1
a) \(-\frac{1}{3}xy\).(3\(x^2yz^2\))
=\(\left(-\frac{1}{3}.3\right)\).\(\left(x.x^2\right)\).(y.y).\(z^2\)
=\(-x^3\).\(y^2z^2\)
b)-54\(y^2\).b.x
=(-54.b).\(y^2x\)
=-54b\(y^2x\)
c) -2.\(x^2y.\left(\frac{1}{2}\right)^2.x.\left(y^2.x\right)^3\)
=\(-2x^2y.\frac{1}{4}.x.y^6.x^3\)
=\(\left(-2.\frac{1}{4}\right).\left(x^2.x.x^3\right).\left(y.y^2\right)\)
=\(\frac{-1}{2}x^6y^3\)
Bài 3:
a) \(f\left(x\right)=-15x^2+5x^4-4x^2+8x^2-9x^3-x^4+15-7x^3\)
\(f\left(x\right)=\left(5x^4-x^4\right)-\left(9x^3+7x^3\right)-\left(15x^2+4x^2-8x^2\right)+15\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
b)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=-8\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(-1\right)=4\cdot\left(-1\right)^4-16\cdot\left(-1\right)^3-11\cdot\left(-1\right)^2+15\)
\(f\left(-1\right)=24\)
Bài 5:
a: \(P\left(x\right)=3x^5+x^4-2x^2+2x\)
\(Q\left(x\right)=-3x^5+2x^2-2x+3\)
b: \(P\left(x\right)+Q\left(x\right)=3x^5-3x^5+x^4-2x^2+2x^2+2x-2x+3\)
\(=x^4+3\)
\(P\left(x\right)-Q\left(x\right)=3x^5+x^4-2x^2+2x+3x^5-2x^2+2x-3\)
\(=6x^5+x^4-4x^2+4x-3\)
c: \(P\left(0\right)=3\cdot0^5+0^4-2\cdot0^2+2\cdot0=2\)
\(Q\left(0\right)=-3\cdot0^5+2\cdot0^2-2\cdot0+3=3\)
Vậy: x=0 là nghiệm của P(x), không là nghiệm của Q(x)
\(M\left(x\right)=P\left(x\right)+Q\left(x\right)=2,5x^6-4+2,5x^5-6x^3+2x^2\)-5x+\(3x-2,5x^6-x^2+5-2,5x^5+6x^3\)
=\(\left(2,5x^6-2,5x^6\right)\)+\(\left(2,5x^5-2,5x^5\right)\)\(\left(-6x^3+6x^3\right)\)+\(\left(2x^2-x^2\right)\)+\(\left(-5x+3x\right)\)+(-4+5)
= \(x^2-2x+1\)
a: \(A\left(x\right)=3x^5+2x^4-4x^3+x^2-2x+1\)
\(=3x^5+2x^4-4x^3+x^2-2x+1\)
\(B\left(x\right)=-x^4+3x^3-2x^2+x^3-3x+2-3x^4\)
\(=\left(-x^4-3x^4\right)+\left(3x^3+x^3\right)+\left(-2x^2-3x+2\right)\)
\(=-4x^4+4x^3-2x^2-3x+2\)
b:
A(x)+B(x)
\(=3x^5+2x^4-4x^3+x^2-2x+1-4x^4+4x^3-2x^2-3x+2\)
\(=3x^5-2x^4-x^2-5x+3\)
A(x)-B(x)
\(=3x^5+2x^4-4x^3+x^2-2x+1+4x^4-4x^3+2x^2+3x-2\)
\(=3x^5+6x^5-8x^3+3x^2+x-1\)
a: \(A \left(\right. x \left.\right) = 3 x^{5} + 2 x^{4} - 4 x^{3} + x^{2} - 2 x + 1\)
\(= 3 x^{5} + 2 x^{4} - 4 x^{3} + x^{2} - 2 x + 1\)
\(B \left(\right. x \left.\right) = - x^{4} + 3 x^{3} - 2 x^{2} + x^{3} - 3 x + 2 - 3 x^{4}\)
\(= \left(\right. - x^{4} - 3 x^{4} \left.\right) + \left(\right. 3 x^{3} + x^{3} \left.\right) + \left(\right. - 2 x^{2} - 3 x + 2 \left.\right)\)
\(= - 4 x^{4} + 4 x^{3} - 2 x^{2} - 3 x + 2\)
b:
A(x)+B(x)
\(= 3 x^{5} + 2 x^{4} - 4 x^{3} + x^{2} - 2 x + 1 - 4 x^{4} + 4 x^{3} - 2 x^{2} - 3 x + 2\)
\(= 3 x^{5} - 2 x^{4} - x^{2} - 5 x + 3\)
A(x)-B(x)
\(= 3 x^{5} + 2 x^{4} - 4 x^{3} + x^{2} - 2 x + 1 + 4 x^{4} - 4 x^{3} + 2 x^{2} + 3 x - 2\)
\(= 3 x^{5} + 6 x^{5} - 8 x^{3} + 3 x^{2} + x - 1\)