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16 tháng 5 2022
a: \(=\dfrac{x^4+15x+7}{x^4+15x+7}\cdot\dfrac{x}{14x^2+1}\cdot\dfrac{4x^3+4}{2x^3+2}=\dfrac{2x}{14x^2+1}\)
b: \(=\dfrac{x^7+3x^2+2}{x^7+3x^2+2}\cdot\dfrac{x^2+x+1}{x^3-1}\cdot\dfrac{3x}{x+1}\)
\(=\dfrac{1}{x-1}\cdot\dfrac{3x}{x+1}=\dfrac{3x}{x^2-1}\)
12 tháng 7 2016
a) \(x^4+3x^3-7x^2-27x-18\)
\(=\left(x^4+3x^3+2x^2\right)-\left(9x^2-27x-18\right)\)
\(=x^2\left(x^2+3x+2\right)-9\left(x^2+3x+2\right)=\left(x^2+x+2x+2\right)\left(x^2-9\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)
b) \(x^4+5x^3-7x^2-41x-30\)
\(=\left(x^4+2x^3-15x^2\right)+\left(3x^3+6x^2-45x\right)+\left(2x^2+4x-30\right)\)
\(=x^2\left(x^2+2x-15\right)+3x\left(x^2+2x-15\right)+2\left(x^2+2x-15\right)\)
\(=\left(x^2+2x-15\right)\left(x^2+3x+2\right)=\left(x^2+5x-3x-15\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+5\right)\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c) \(x^6-14x^4+49x^2-36\)
\(=\left(x^6-9x^4\right)+\left(-5x^4+45x^2\right)+\left(4x^2-36\right)\)
\(=x^4\left(x^2-9\right)-5x^2\left(x^2-9\right)+4\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(x^4-5x^2+4\right)=\left(x^2-9\right)\left(x^4-4x^2-x^2+4\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)
10 tháng 7 2017
Theo đề bài ta có :
\(\frac{x\left(3-x\right)}{x+1}\cdot\left(x+\frac{\left(3-x\right)}{x+1}\right)=2\)
=> \(\frac{\left(3x-x^2\right)}{x+1}\cdot\frac{\left(3-x+x^2+x\right)}{x+1}=2\)
=> \(\left(3x-x^2\right)\left(x^2+3\right)=2\left(x+1\right)^2\)
=> \(3x^3+9x-x^4-3x^2=2x^2+4x+2\)
=> \(3x^3+\left(9x-4x\right)+\left(-3x^2-2x^2\right)-x^4-2=0\)
=> \(3x^3+5x-5x^2-x^4-2=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x^3-1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)-2\left(1-x\right)\left(x^2+x+1\right)=0\)
=> \(\left(1-x\right)\left(5x+x^3-2x^2-2x-2\right)=0\)
=> \(\left(1-x\right)\left(3x+x^3-2x^2-2\right)=0\)
=> \(\left(1-x\right)\left(x^3-x^2-x^2+x+2x-2\right)=0\)
=> \(\left(1-x\right)\left(x^2\left(x-1\right)-x\left(x-1\right)+2\left(x-1\right)\right)=0\)
=> \(\left(1-x\right)\left(x-1\right)\left(x^2-x+2\right)=0\)
Ta Thấy :
\(\left(x^2-x+2\right)=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=> \(\hept{\begin{cases}1-x=0\\x-1=0\end{cases}}\)
=> x = 1
NT
24 tháng 12 2017
a)
\(\dfrac{36\left(x-2\right)^3}{32-16x}=\dfrac{9\left(x-2\right)^3}{8-4x}=\dfrac{9\left(x-2\right)^3}{4\left(2-x\right)}=\dfrac{9\left(x-2\right)^3}{-4\left(x-2\right)}=\dfrac{9\left(x-2\right)^2}{-4}\)
b)
\(\dfrac{x^2+2x+1}{x+1}=\dfrac{\left(x+1\right)^2}{x+1}=x+1\)
c)
\(\dfrac{x^2-2x+1}{x^2-1}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
29 tháng 6 2017
a, \(\dfrac{3\left|x-4\right|}{3x^2-3x-36}=\dfrac{3\left|x-4\right|}{3\left(x^2-x-12\right)}=\dfrac{\left|x-4\right|}{x^2-x-12}\)
b, \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}=\dfrac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)
\(=\dfrac{\left(-x-2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(x+8\right)}{x+2}=\dfrac{-x-8}{x+2}\)
c, \(\dfrac{x^2+5x+6}{x^2+4x+4}=1+\dfrac{x+2}{x^2+4x+4}=1+\dfrac{x+2}{\left(x+2\right)^2}\)
\(=1+\dfrac{1}{x+2}=\dfrac{x+3}{x+2}\)
29 tháng 6 2017
a, tiếp:
+) Xét \(x\ge4\) có:
\(\dfrac{x-4}{x^2-x-12}=\dfrac{x-4}{x^2-4x+3x-12}=\dfrac{x-4}{x\left(x-4\right)+3\left(x-4\right)}\)
\(=\dfrac{x-4}{\left(x+3\right)\left(x-4\right)}=\dfrac{1}{x+3}\)
+) Xét x < 4 có:
\(\dfrac{4-x}{x^2-x-12}=\dfrac{4-x}{x^2-4x+3x-12}=\dfrac{4-x}{x\left(x-4\right)+3\left(x-4\right)}\)
\(=\dfrac{4-x}{\left(x+3\right)\left(x-4\right)}=\dfrac{-\left(x-4\right)}{\left(x+3\right)\left(x-4\right)}=\dfrac{-1}{x+3}\)
6 tháng 2 2018
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn