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xét tính cẵn lẻ của các hàm số sau a.y=x^4-4x ²+2 b.y=-2x ³+3x c.y=I x=2 I – I x-2 I d.y=I 2x+1 I + I2x-1 I e.y=(x-1) ² f.y=x ²+x
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Đáp án:
$\begin{array}{l}
a)f\left( x \right) = y = {x^4} – 4{x^2} + 2\\
\Rightarrow f\left( { – x} \right) = {\left( { – x} \right)^4} – 4{\left( { – x} \right)^2} + 2\\
= {x^4} – 4{x^2} + 2\\
\Rightarrow f\left( { – x} \right) = f\left( x \right)\\
\Rightarrow hs\,\text{chẵn}\\
b)f\left( x \right) = – 2{x^3} + 3x\\
\Rightarrow f\left( { – x} \right) = – 2.{\left( { – x} \right)^3} + 3.\left( { – x} \right)\\
= 2{x^3} – 3x\\
\Rightarrow f\left( { – x} \right) = – f\left( x \right)\\
\Rightarrow hs\,\text{lẻ}\\
c)y = \left| {x + 2} \right| – \left| {x – 2} \right|\\
\Rightarrow f\left( { – x} \right) = \left| { – x + 2} \right| – \left| { – x – 2} \right|\\
= \left| {x – 2} \right| – \left| {x + 2} \right|\\
\Rightarrow f\left( { – x} \right) = – f\left( x \right)\\
\Rightarrow hs\,\text{lẻ}\\
d)y = \left| {2x + 1} \right| + \left| {2x – 1} \right|\\
\Rightarrow f\left( { – x} \right) = \left| { – 2x + 1} \right| + \left| { – 2x – 1} \right|\\
= \left| {2x – 1} \right| + \left| {2x + 1} \right|\\
\Rightarrow f\left( { – x} \right) = f\left( x \right)\\
\Rightarrow hs\,\text{chẵn}\\
e)f\left( x \right) = {\left( {x – 1} \right)^2}\\
\Rightarrow f\left( { – x} \right) = {\left( { – x – 1} \right)^2} = {\left( {x + 1} \right)^2}\\
\Rightarrow f\left( x \right) \ne f\left( { – x} \right)\\
\Rightarrow hs\,ko\,\text{chẵn};ko\,\text{lẻ}\\
f)y = f\left( x \right) = {x^2} + x\\
\Rightarrow f\left( { – x} \right) = {\left( { – x} \right)^2} – x = {x^2} – x \ne f\left( x \right)\\
\Rightarrow hs\,ko\,\text{chẵn},ko\,\text{lẻ}
\end{array}$