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tìm a,b,c biết rằng P(x) = x^3 + ax^2 +bx + c chia hết cho x-2 và chia x^2 -1 dư 2x
GIÚP MÌNH NHA!...
\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x^2-3x-2x+6}+\frac{1}{x^2-3x-4x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-5x-6x+30}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{1-5}-\frac{1}{1-4}+\frac{1}{1-4}-\frac{1}{1-3}+\frac{1}{1-3}-\frac{1}{1-2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{x^2-8x+12}=\frac{1}{8}\)
\(\Leftrightarrow x^2-8x+12=32\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow x=10\) hoặc \(x=-2\)
\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)\(\frac{1}{x^2-2x-3x+6}+\frac{1}{x^2-4x-3x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-6x-5x+30}=\frac{1}{8}\)
\(\frac{1}{x\left(x-2\right)-3\left(x-2\right)}+\frac{1}{x\left(x-4\right)-3\left(x-4\right)}+\frac{1}{x\left(x-4\right)-5\left(x-4\right)}+\frac{1}{x\left(x-6\right)-5\left(x-6\right)}=\frac{1}{8}\)
\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)dhjjhhjhhjj
\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)
Còn lại tự giải quyết nha
a) Ta có : \(f\left(x\right)+3f\left(\frac{1}{3}\right)=x^2\left(1\right)\Rightarrow f\left(\frac{1}{3}\right)+3f\left(\frac{1}{3}\right)=\left(\frac{1}{3}\right)^2\Leftrightarrow4f\left(\frac{1}{3}\right)=\frac{1}{9}\Leftrightarrow f\left(\frac{1}{3}\right)=\frac{1}{36}\)
Thay f(\(\frac{1}{3}\)) = \(\frac{1}{36}\) vào (1) được : \(f\left(x\right)=x^2-3f\left(\frac{1}{3}\right)=x^2-\frac{1}{12}\)
Vậy \(f\left(x\right)=x^2-\frac{1}{12}\)
b) \(f\left(x\right)+2f\left(\frac{1}{x}\right)=2x+\frac{1}{x}\) (2) . Thay \(x=\frac{1}{x}\) vào \(f\left(x\right)\) và \(f\left(\frac{1}{x}\right)\) được :
\(f\left(\frac{1}{x}\right)+2f\left(x\right)=\frac{2}{x}+x\) \(\Leftrightarrow2f\left(\frac{1}{x}\right)+4f\left(x\right)=\frac{4}{x}+2x\) (3)
Lấy (3) trừ (2) theo vế được: \(\left[2f\left(\frac{1}{x}\right)+4f\left(x\right)\right]-\left[f\left(x\right)+2f\left(\frac{1}{x}\right)\right]=\left(2x+\frac{4}{x}\right)-\left(2x+\frac{1}{x}\right)\)
\(\Leftrightarrow3f\left(x\right)=\frac{3}{x}\Leftrightarrow f\left(x\right)=\frac{1}{x}\)
c) \(f\left(x\right)+2f\left(-x\right)=x+1\) (4) . Thay x = -x vào f(x) và f(-x) được :
\(f\left(-x\right)+2f\left(x\right)=-x+1\Leftrightarrow2f\left(-x\right)+4f\left(x\right)=-2x+2\) (5)
Lấy (5) trừ (4) theo vế được :
\(\left[2f\left(-x\right)+4f\left(x\right)\right]-\left[f\left(x\right)+2f\left(-x\right)\right]=\left(-2x+2\right)-\left(x+1\right)\)
\(\Leftrightarrow3f\left(x\right)=-3x+1\Rightarrow f\left(x\right)=\frac{-3x+1}{3}\)