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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
1)\(4\left(a^4-1\right)x=5\left(a-1\right)\)
<=>x=\(\frac{5\left(a-1\right)}{a^4-1}\)
<=>x=\(\frac{5\left(a-1\right)}{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)}=\frac{5}{\left(a+1\right)\left(a^2+1\right)}\)
Tương tự ta tính được y=\(\frac{4a^6+4}{5a^4-5a^2+5}\)
Suy ra x.y=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\cdot\left(a^6+1\right)}{5\left(a^4-a^2+1\right)}\)=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\left(a^2+1\right)\left(a^4-a^2+1\right)}{5\left(a^4-a^2+1\right)}\)
=\(\frac{5}{a+1}\)
Tương tự với x:y
\(A=\frac{4.6}{4.2}:\left(\frac{8.10}{6.8}.\frac{12.14}{10.12}.\frac{16.18}{14.16}...\frac{54.56}{54.53}\right)=\frac{6}{2}:\frac{56}{6}=\)
a: \(A=25a^2+50a+25+10\left(a^2-2a-3\right)+a^2-6a+9\)
\(=26a^2+46a+34+10a^2-20a-30\)
\(=36a^2+26a+4\)
b: \(B=\dfrac{1}{4}\left(x^2-2x+1\right)+x^2-1+x^2+2x+1\)
\(=\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}+2x^2+2x\)
\(=\dfrac{9}{4}x^2+\dfrac{3}{2}x+\dfrac{1}{4}\)
1, \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right)\left(4x+3\right)\)
2,\(x^2-4+\left(x+2\right)^2=\left(x-2\right)\left(x+2\right)\left(x+2\right)^2=\left(x-2\right)\left(x+2\right)^3\)
3,\(5a\left(a-2\right)-a+2=5a\left(a-2\right)-1\left(a-2\right)=\left(5a-1\right)\left(a-2\right)\)
4,\(7\left(a-5\right)+8a\left(5-a\right)=7\left(a-5\right)-8a\left(a-5\right)=\left(7-8a\right)\left(a-5\right)\)
5, \(25a^2-4b^2+4b-1=25a^2-\left(4b^2-4b+1\right)=\left(5a\right)^2-\left(2b-1\right)^2=\left(5a-2b+1\right)\left(5a+2b-1\right)\)