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\(\frac{a}{x-2}+\frac{b}{\left(x+1\right)^2}=\frac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)
\(\Rightarrow\frac{x^2+5}{x^3-3x-2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)
Đồng nhất hệ số, ta có :
\(\hept{\begin{cases}a=1\\2a+b=0\\a-2b=5\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=-2\end{cases}}}\)
\(x^4-4x^3+4x^2\)
\(=x^2\left(x^2-4x+4\right)\)
\(=x^2\left(x-2\right)^2\)
\(3x^2+10x+3\)
\(=3x^2+x+9x+3\)
\(=x\left(3x+1\right)+3\left(3x+1\right)\)
\(=\left(x+3\right)\left(3x+1\right)\)
\(x^4-4x^3+4x^2\)
\(=x^2.\left(x^2-2.x.2+2^2\right)\)
\(=x^2.\left(x-2\right)^2\)
\(\dfrac{x^3-x^2-x+1}{x^4-2x^2+1}=\dfrac{x^2\left(x-1\right)-\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}=\dfrac{1}{x+1}\)
\(\dfrac{5x^3+10x^2+5x}{x^3+3x^2+3x+1}=\dfrac{5x\left(x+1\right)^2}{\left(x+1\right)^3}=\dfrac{5x}{x+1}\)
\(x^4-4x^3+9x^2-10x+6\)
\(=x^4-2x^3+3x^2-2x^3+4x^2-6x+2x^2-4x+6\)
\(=x^2\left(x^2-2x+3\right)-2x\left(x^2-2x+3\right)+2\left(x^2-2x+3\right)\)
\(=\left(x^2-2x+3\right)\left(x^2-2x+2\right)\)
- x2.(x3-x2+x-1)
- x.( x3-3x2-1)+3
- x.(x2-xy-y2)
Tìm x:
x3-16x = 0
=> x.(x2-16) = 0
=> x = 0 hay x2-16 = 0
=> x = 0 hay x2 = 0+16
=> x = 0 hay x2 = 16
=> x = 0 hay x = 4 hay x = -4