Cho hai đa thức \(f\left(x\right)=x^5-3x^2+7x^4-9x^3+x^2-\frac{1}{4}x\)
\(g\left(x\right)=5x^4-x^5+x^2-2x^3+3x^2-\frac{1}{4}\)
Tính \(f\left(x\right)+g\left(x\right)\)và \(f\left(x\right)-g\left(x\right)\)
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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
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)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=-8\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(-1\right)=4\cdot\left(-1\right)^4-16\cdot\left(-1\right)^3-11\cdot\left(-1\right)^2+15\)
\(f\left(-1\right)=24\)
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) A(x) = f(x) + g(x) = ( 2x^3 + 3x - 4x^3 + 1/2 - 5x^4 ) + ( 3x^4 + 0,2 - 7x^2 + 5x^3 - 9x )
= 2x^3 + 3x - 4x^3 + 1/2 - 5x^4 + 3x^4 + 0,2 - 7x^2 + 5x^3 - 9x
= ( 2x^3 - 4x^3 + 5x^3 ) + ( 3x - 9x ) + ( 1/2 + 0,2 ) + ( -5x^4 + 3x^4 ) - 7x^2
= 3x^3 - 6x + 0,7 - 2x^4 - 7x^2
B(x) = f(x) - g(x) = ( 2x^3 + 3x - 4x^3 + 1/2 - 5x^4 ) - ( 3x^4 + 0,2 - 7x^2 + 5x^3 - 9x )
= 2x^3 + 3x - 4x^3 + 1/2 - 5x^4 - 3x^4 - 0,2 + 7x^2 - 5x^3 + 9x
= ( 2x^3 - 4x^3 - 5x^3 ) + ( 3x + 9x ) + ( 1/2 - 0,2 ) + ( -5x^4 - 3x^4 ) + 7x^2
= -7x^3 + 12x + 0,3 -8x^4 + 7x^2
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)