Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
a: \(\dfrac{7x^3y^4}{35xy}=\dfrac{7xy\cdot x^2y^3}{7xy\cdot5}=\dfrac{x^2y^3}{5}\)
b: \(\dfrac{x^3-4x}{10-5x}=\dfrac{-x\left(x-2\right)\left(x+2\right)}{5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}\)
c: \(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+2}{x-1}\)
d: \(\left(x^2-x-2\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x^2-3x+2\right)\left(x+1\right)\)
=>\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)
e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)
Bài 1.
a) \(\left(x^5+4x^3-6x^2\right):4x^2\)
\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)
\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
b)
Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)
c) (-2x5 : 2x2) + (3x2 : 2x2) + (-4x^3 : 2x^2)
= \(-x^3+\dfrac{3}{2}-2x\)
d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)
\(=x-4\)
(dùng hẳng đẳng thức thứ 7)
Bài 2 :
a) 3x(x - 2) - 5x(1 - x) - 8(x2 - 3)
= 3x2 - 6x - 5x + 5x2 - 8x2 + 24
= (3x2 + 5x2 - 8x2) + (-6x - 5x) + 24
= -11x + 24
b) (x - y)(x2 + xy + y2) + 2y3
= x3 - y3 + 2y3
= x3 + y3
c) (x - y)2 + (x + y)2 - 2(x - y)(x + y)
= (x - y)2 - 2(x - y)(x + y) + (x + y)2
= [(x - y) + x + y)2 = [x - y + x + y] = (2x)2 = 4x2
Bài 1 :
a]= \(\frac{1}{4}\)x3 + x - \(\frac{3}{2}\).
b] => [x3 + x2 -12 ] = [ x2 +3 ][x-2] + [-6]
c]= -x3 -2x +\(\frac{3}{2}\).
d] = [ x3 - 64 ] = [ x2 + 4x + 16][ x- 4].
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
a) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
b) \(\dfrac{9}{x^3-9x}-\dfrac{-1}{x+3}\)
\(=\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\)
c) \(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}\)
\(=\dfrac{x\left(x-2\right)\left(x^2+2x+4\right)\left(x+4\right)}{5\left(x+2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{x\left(x-2\right)\left(x+4\right)}{5\left(x+2\right)}\)
d) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{2\left(2-x\right)}{x+2}\)
\(=-\dfrac{10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{5}{2}\)
e) \(\dfrac{\left(x-13\right)^2}{2x^5}.\dfrac{-3x^2}{x-13}\)
\(=\dfrac{x-13}{2x^3}.\dfrac{-3}{1}\)
\(=\dfrac{-3\left(x-13\right)}{2x^3}\)
g) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2\left(x-1\right)^2}{2\left(x-1\right)\left(x+3\right)^2}\)
\(=-\dfrac{x-1}{2}\).