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(x^4-27x):(x^2+3x+9)
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9 tháng 10 2021
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
23 tháng 3 2022
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
14 tháng 8 2021
a: \(\dfrac{x^2}{3x+6}+\dfrac{4x+4}{3x+6}=\dfrac{x^2+4x+4}{3x+6}=\dfrac{x+2}{3}\)
b: \(\dfrac{x+3}{x}+\dfrac{x}{3-x}-\dfrac{9}{3x-x^2}\)
\(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}\)
=0
23 tháng 12 2021
a: \(=x^2+2x-8-x^2-2x-1=-9\)
b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
17 tháng 8 2021
h, \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(\Rightarrow\left(27x^3-8\right):\left(3x-2\right)\\ =\left(3x-2\right)\left(9x^2+6x+4\right):\left(3x-2\right)\\ =9x^2+6x+4\)
g, \(x^4-2x^2+1=\left(x^2-1\right)^2\)
\(\Rightarrow\left(x^4-2x^2+1\right):\left(1-x^2\right)\\ =\left(x^2-1\right)^2:\left(1-x^2\right)\\ =x^2-1\)
TK
1
Q
5
PN
8 tháng 8 2020
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)
\(=x^3-8-\left(x^3-3x^2+3x-1\right)+7\)
\(=x^3-8+7-x^3+3x^2-3x+1\)
\(=\left(x^3-x^3\right)+\left(7+1-8\right)+3x^2-3x\)
\(=3x^2-3x=3x\left(x-1\right)\)
PN
8 tháng 8 2020
\(x\left(x+2\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=x\left(2+x\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=x\left(4-x^2\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=4x-x^3+\left(x^3+9\right)\)
\(=4x-\left(x^3-x^3\right)+9\)
\(=4x+9\)
BT
0
2 tháng 12 2023
Bài 4:
1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)
=>\(x^3-1-x^3-6x=11\)
=>-6x-1=11
=>-6x=11+1=12
=>\(x=\dfrac{12}{-6}=-2\)
2: \(16x^2-\left(3x-4\right)^2=0\)
=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)
=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)
=>(x+4)(7x-4)=0
=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)
3: \(x^3-x^2-3x+3=0\)
=>\(\left(x^3-x^2\right)-\left(3x-3\right)=0\)
=>\(x^2\left(x-1\right)-3\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-3\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))
=>\(\left(x+2\right)^2=\left(x-1\right)\left(x+1\right)\)
=>\(x^2+4x+4=x^2-1\)
=>4x+4=-1
=>4x=-5
=>\(x=-\dfrac{5}{4}\left(nhận\right)\)
5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)
=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)
=>3x+1=0
=>3x=-1
=>\(x=-\dfrac{1}{3}\left(nhận\right)\)
6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)
\(\dfrac{9-x^2}{x}:\left(x-3\right)=1\)
=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-x-3}{x}=1\)
=>-x-3=x
=>-2x=3
=>\(x=-\dfrac{3}{2}\left(nhận\right)\)
Answer:
\(\left(x^4-27x\right):\left(x^2+3x+9\right)\)
\(=[x\left(x^3-27\right)]:\left(x^2+3x+9\right)\)
\(=[x\left(x-3\right)\left(x^2+3x+9\right)]:\left(x^2+3x+9\right)\)
\(=x\left(x-3\right)\)
\(=x^2-3x\)