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Ta có:
\(a^5-a^4-18a^3+9a^2-5a+2017+\frac{a^4-40a^2+4}{a^2}\)
\(=a^5-5a^4+2a^3+4a^4-20a^3+8a^2+a^2-5a+2+2015+\frac{a^4-40a^2+4}{a^2}\)
\(=\left(a^2-5a+2\right)\left(a^3+4a^2+1\right)+2015+\frac{a^4-40a^2+4}{a^2}\)
\(=2015+\frac{a^4-40a^2+4}{a^2}=\frac{a^4+1970a^2+4}{a^2}\)
\(a^2-5a+2=0\Rightarrow a^2-5a=-2\Rightarrow a^4-10a^3+25a^2=4\)
Ta có : \(\frac{a^4+1970a^2+4}{a^2}=\frac{a^4-10a^3+25a^2+10a^3-50a^2+20a+4a^2-20a+8+1991a^2-4}{a^2}\)
\(=\frac{4+\left(10a+4\right)\left(a^2-5a+2\right)-4+1991a^2}{a^2}\)
\(=\frac{1991a^2}{a^2}=1991\)
a) \(ĐKXĐ:x\ne\pm3\)
\(A=\frac{5}{x+3}-\frac{2}{3-x}+\frac{3x^2-2x-9}{x^2-9}\)
\(\Leftrightarrow A=\frac{5\left(x-3\right)+2\left(x+3\right)-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x}{x+3}\)
b) Khi \(\left|x-2\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\2-x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Thay x = 1 vào A, ta được :
\(A=\frac{-3}{1+3}=\frac{-3}{4}\)
Vậy khi \(\left|x-2\right|=1\Leftrightarrow A=-\frac{3}{4}\)
c) Để \(A\inℤ\)
\(\Leftrightarrow\frac{-3x}{x+3}\inℤ\)
\(\Leftrightarrow-3x⋮x+3\)
\(\Leftrightarrow-3\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow9⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)
Ta có:\(a^2-5a+2=0\Rightarrow a^2=5a-2\)
\(P=a^5-a^4-18a^3+9a^2-5a+2017+\frac{a^4-40a^2+4}{a^2}\)
\(=a^5-a^4-18a^3+9a^2-5a+2017+\frac{\left(a^2-2\right)^2-36a^2}{a^2}\)
\(=a^5-a^4-18a^3+9a^2-5a+2015+2+\frac{\left(a^2-2\right)^2-\left(6a\right)^2}{a^2}\)
\(=\left(a^2-5a+2\right)\left(a^3+4a^2+1\right)+2015+\frac{\left(a^2-2+6a\right)\left(a^2-2-6a\right)}{a^2}\)
\(=0\times\left(a^3+4a^2+1\right)+2015+\frac{\left(a^2-2+6a\right)\left(a^2-2-6a\right)}{a^2}\)
\(=0+2015+\frac{\left(a^2-2+6a\right)\left(a^2-2-6a\right)}{a^2}\)
\(=2015+\frac{\left(5a-2-6a-2\right)\left(5a-2+6a-2\right)}{a^2}\)Vì \(a^2=5a-2\)
\(=2015+\frac{-\left(a+4\right)\left(11a-4\right)}{a^2}\)
\(=2015+\frac{-\left(a^2+40a-16\right)}{a^2}\)
\(=2015+\frac{-\left[a^2+8\left(5a-2\right)\right]}{a^2}\)Vì \(a^2=5a-2\)
\(=2015+\frac{-\left(a^2+8a^2\right)}{a^2}\)
\(=2015+\frac{-9a^2}{a^2}\)
\(=2015+\frac{-9}{1}\)
\(=2015-9\)
\(=2006\)
Cre:hoidap247