Thực hiện phép chia sau :
a,\(x^6+x^7-x^9:x\) g,\(\left(x^2+5x-6\right):\left(x-1\right)\)
b, \(6x^4+4x^3+8x^2:2x^3\) f, \(\left(x^2+4x-3\right):\left(x-3\right)\)
c,\(14x^4+21x^5-7x^7:x^3\) h, \(\left(x^2+4x-7\right):\left(x-2\right)\)
d,\(\left(-x^2+4x\right):\left(x-4\right)\) i,\(\left(x^3+5x^2+11x+10\right):\left(x+2\right)\)
e,\(\left(x^2+x-12\right):\left(x-3\right)\)
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a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)
c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)
\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
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:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
a) Thu gọn, sắp xếp các đa thức theo lũy thừa tăng của biến
= -9 - 2x2 + 3x3 - 6x5 - 3x7
b) Tính -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
\(a)\left(x^6+x^7-x^9\right):x\\ =x^6:x+x^7:x-x^9:x\\ =x^5+x^6x^8\\ b)\left(6x^4+4x^3+8x^2\right):2x^2\\ =6x^4:2x^2+4x^3:2x^2+8x^2:2x^2\\ =3x^2+2x+4\\ c)\left(14x^4+21x^5+7x^7\right):x^3\\ =14x^4:x^3+21x^5:x^3+7x^7:x^3\\ =14x+21x^2+7x^4\\ d)\left(-x^2+4x\right):\left(x-4\right)\\ =-x\left(x-4\right):\left(x-4\right)\\ =-x\)
\(e)\left(x^2+x-12\right):\left(x-3\right)\\ =\left(x^2-3x+4x-12\right):\left(x-3\right)\\ =\left[x\left(x-3\right)+4\left(x-3\right)\right]:\left(x-3\right)\\ =\left(x-3\right)\left(x+4\right):\left(x-3\right)\\ =x+4\\ g)\left(x^2+5x-6\right):\left(x-1\right)\\ =\left(x^2-x+6x-6\right):\left(x-1\right)\\ =\left[x\left(x-1\right)+6\left(x-1\right)\right]:\left(x-1\right)\\ =\left(x+6\right)\left(x-1\right):\left(x-1\right)\\ =x+6\\ i)\left(x^3+5x^2+11x+10\right):\left(x+2\right)\\ =\left(x^3+2x^2+3x^2+6x+5x+10\right):\left(x+2\right)\\ =\left[x^2\left(x+2\right)+3x\left(x+2\right)+5\left(x+2\right)\right]:\left(x+2\right)\\ =\left(x^2+3x+5\right)\left(x+2\right):\left(x+2\right)\\ =x^2+3x+5\)
(xem lại đề câu f vs h)