a) Thu gọn, sắp xếp các đa thức theo lũy thừa tăng của biến

$f\left(x\right)=x^2+2x^3-7x^5-9-6x^7+x^3+x^2+x^5-4x^2+3x^7$

= -9 - 2x² + 3x³ - 6x⁵ - 3x⁷

$g\left(x\right)=x^5+2x^3-5x^8-x^7+x^3+4x^2-5x^7+x^4-4x^2-x^6-12$

$\mathrm{=\; -12\; +\; 3x3\; +\; x4\; +\; x5\; -\; x6\; -\; 6x7\; -\; 5x8}$

$h\left(x\right)=x+4x^5-5x^6-x^7+4x^3+x^2-2x^7+x^6-4x^2-7x^7+x$

$\mathrm{=\; 2x\; -\; 3x2\; +\; 4x3\; +4x5\; -4x6\; -\; 10x7}$

b) Tính $f\left(x\right)+g\left(x\right)-h(x)\; =\; ($-9 - 2x² + 3x³ - 6x⁵ - 3x⁷ ) + (-12 + 3x³ + x⁴ + x⁵ - x⁶ - 6x⁷ - 5x⁸ ) - (2x - 3x² + 4x³ +4x⁵ -4x⁶ - 10x⁷)

= - 9 - 2x² + 3x³ - 6x⁵ - 3x⁷ -12 + 3x³ + x⁴ + x⁵ - x⁶ - 6x⁷ - 5x⁸ - 2x + 3x² - 4x³ - 4x⁵ + 4x⁶ + 10x⁷

= -21 - 2x + x² + 2x³ + x⁴ - 9x⁵ + 3x⁶ + x⁷ - 5x⁸

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)=\(-2^4+8.-2^3+4.-2^2+6.-2\)\(-10\)

         =\(-54\)

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:

\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow-6< x< 2\)

5.

\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)

\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)

6.

\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)

\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)

Ta có: A(x) = -4x⁵ - x³ + 4x² + 5x + 9 + 4x⁵ - 6x² - 2

A(x) = (-4x⁵ + 4x⁵) - x³ + (4x² - 6x²) + 5x + (9 - 2)

A(x) = -x³ - 2x² + 5x + 7

B(x) = -3x⁴ - 2x³ + 10x² - 8x + 5x³ - 7 - 2x³ + 8x

B(x) = -3x⁴ - (2x³ - 5x³ + 2x³) + 10x² - (8x - 8x) - 7

B(x) = -3x⁴ + x³ + 10x² - 7

A(x) + B(x) = (-x³ - 2x² + 5x + 7) + (-3x⁴ + x³ + 10x² - 7)

  = -x³ - 2x² + 5x + 7 - 3x⁴ + x³ + 10x² - 7

 = (-x³ + x³) - (2x² - 10x²) + 5x + (7 - 7)

 = 8x² + 5x

A(x) - B(x) = (-x^3 - 2x^2 + 5x + 7) - (-3x^4 + x^3 + 10x^2 - 7)

= -x^3 - 2x^2 + 5x + 7 + 3x^4 - x^3 - 10x^2 + 7

= (-x^3 - x^3) - (2x^2 + 10x^2) + 5x + (7 + 7)

= -2x^3 - 12x^2 + 5x + 14

\(A\left(x\right)=-x^3-2x^2+5x+7\)

\(B\left(x\right)=-3x^4=x^3+10x^2-7\)

\(A\left(x\right)+B\left(x\right)=8x^2+5x\)

\(A\left(x\right)-B\left(x\right)=-2x^3-12x^2+5x+14\)

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) \(A\left(x\right)=3x^3-4x^4-2x^3+4x^4-5x+3\)

\(\Rightarrow A\left(x\right)=-4x^4+4x^4+3x^3-2x^3-5x+3\)

\(\Rightarrow A\left(x\right)=x^3-5x+3\)

\(B\left(x\right)=5x^3-4x^2-5x^3-4x^2-5x-3\)

\(\Rightarrow B\left(x\right)=5x^3-5x^3-4x^2-4x^2-5x-3\)

\(\Rightarrow B\left(x\right)=-8x^2-5x-3\)

b) \(A\left(x\right)+B\left(x\right)=x^3-5x+3+\left(-8x^2-5x-3\right)\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-5x-5x+3-3\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\)

\(A\left(x\right)-B\left(x\right)=x^3-5x+3-\left(-8x^2-5x-3\right)\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3-5x+3+8x^2+5x+3\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2-5x+5x+3+3\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2+6\)

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 1:

a) \(-\frac{1}{3}xy\cdot\left(3x^2yz^2\right)\)

\(=\left(-\frac{1}{3}\cdot3\right)\left(xx^2\right)\left(yy\right)z\)

\(=-x^3y^2z\)

b) \(-54y^2\cdot bx\)

\(=\left(-54b\right)xy^2\)

c) \(-2x^2y\cdot\left(\frac{1}{2}\right)^2\cdot x\cdot\left(y^2x\right)^3\)

\(=-2x^2y\cdot\frac{1}{4}\cdot x\cdot y^5x^3\)

\(=\left(-2\cdot\frac{1}{4}\right)\left(x^2xx^3\right)\left(yy^5\right)\)

\(=-\frac{1}{2}x^6y^6\)

a,

C(x)=-3x^4-2x^3+x^2+x+5