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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)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=-8\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(-1\right)=4\cdot\left(-1\right)^4-16\cdot\left(-1\right)^3-11\cdot\left(-1\right)^2+15\)
\(f\left(-1\right)=24\)
Nguyễn Ngọc Quý bạn tl đi nhé ^^, r bạn Bùi tích đúng cho ^^
\(a,\)
\(\Rightarrow f\left(x\right)=x^4-x^3+3x-1\)
\(\Rightarrow g\left(x\right)=x^4+4x^3+x-5\)
\(b,\)
\(A\left(x\right)=f\left(x\right)-g\left(x\right)=x^4-x^3+3x-1-x^4-4x^3-x+5\)
\(=-5x^3-x+4\)
\(B\left(x\right)=f\left(x\right)+g\left(x\right)=x^4-x^3+3x-1+x^4+4x^3+x-5\)
\(=2x^4+3x^3+4x-6\)
\(c,\)
Thay \(x=-2\) vào \(A\left(x\right)\) , ta được :
\(A\left(x\right)=-5.\left(-2\right)^3+2+4=46\)
Thay \(x=2\) vào \(A\left(x\right)\) , ta được :
\(A\left(x\right)=-5.2^3-2+4=-38\)
f(x) + g(x)
= (x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x) + (5x4 - x5 +x2 - 2x3 + 3x2 - 1/4)
= x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x + 5x4 - x5 +x2 - 2x3 + 3x2 - 1/4
=12x4 - 11x3 + 2x2 - 1/4x - 1/4
f(x) - g(x)
= (x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x) - (5x4 - x5 +x2 - 2x3 + 3x2 - 1/4)
= = x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x - 5x4 + x5 - x2 + 2x3 - 3x2 + 1/4
= 2x5 + 2x4 - 7x3 - 6x2 - 1/4x + 1/4
a)Sắp xếp : \(f\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x\)
\(g\left(x\right)=-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}x\)
Ta có : \(f\left(x\right)+g\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}x\)
\(=12x^4-11x^3+2x^2-\dfrac{1}{2}x\)
\(f\left(x\right)-g\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x+x^5-5x^4+2x^3-4x^2+\dfrac{1}{4}x\)
\(=2x^5+2x^4-7x^3-6x^2\)
Bài 1:
a) Ta có: \(P\left(x\right)=3x^4+2x^2-3x^4-2x^2+2x-5\)
\(=\left(3x^4-3x^4\right)+\left(2x^2-2x^2\right)+2x-5\)
\(=2x-5\)
Bài 1:
b)
\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)
\(P\left(3\right)=2\cdot3-5=6-5=1\)
f(x)+g(x)=12x4-11x3+2x2-\(\frac{1}{4}\)x-\(\frac{1}{4}\)
Con f(x)-g(x) thi tru 2 da thuc tren cho nhau
a, f(x) = -1/4 - 3x2 - 9x3 + 7x4 + x5
g(x) = 2x2 - x5 + 54 - 1/4