Cho hàm số \(y=f\left(x\right)=kx\)(\(k\)là hằng số, \(k\ne0\)). Chứng minh rằng
\(f\left(51x_1-2014x_2\right)=51f\left(x_1\right)-2014f\left(x_2\right)\)
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ta có:
\(f\left(x_1\right)=kx_1;f\left(x_2\right)=kx_2=>f\left(x_1-x_2\right)=k.\left(x_1-x_2\right)=kx_1-kx_2\)
vậy \(f\left(x_1-x_2\right)=f\left(x_1\right)-f\left(x_2\right)\)
tick mk nhé
a) Ta có:
\(f\left( 1 \right) = 1 + 1 = 2\)
\(f\left( 2 \right) = 2 + 1 = 3\)
\( \Rightarrow f\left( 2 \right) > f\left( 1 \right)\)
b) Ta có:
\(f\left( {{x_1}} \right) = {x_1} + 1;f\left( {{x_2}} \right) = {x_2} + 1\)
\(\begin{array}{l}f\left( {{x_1}} \right) - f\left( {{x_2}} \right) = \left( {{x_1} + 1} \right) - \left( {{x_2} + 1} \right)\\ = {x_1} - {x_2} < 0\end{array}\)
Vậy \({x_1} < {x_2} \Rightarrow f\left( {{x_1}} \right) < f\left( {{x_2}} \right)\).
\(f\left(x_1\right)=ax_1\) ; \(f\left(x_2\right)=ax_2\) ; \(f\left(x_1x_2\right)=ax_1x_2\)
Để \(f\left(x_1\right)f\left(x_2\right)=f\left(x_1x_2\right)\)
\(\Leftrightarrow ax_1.ax_2=ax_1x_2\)
\(\Leftrightarrow a^2x_1x_2=ax_1x_2\)
\(\Leftrightarrow a^2=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)
Vậy \(a=1\)
a) theo tính chất ta có: f(0+0)= f(0)+f(0)
=> f(0)=f(0)+f(0)
=> f(0)-f(0)=f(0)+f(0)-f(0)
=> 0=f(0)
hay f(0)=0
b) f(0)=f(-x+x)=f(-x)+f(x)
=>0=f(-x)+f(x)
=> f(-x)=0-f(x)=-f(x)
c) \(f\left(x_1-x_2\right)=f\left(x_1+\left(-x_2\right)\right)=f\left(x_1\right)+f\left(-x_2\right)=f\left(x_1\right)-f\left(x_2\right)\)
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\(f\left(x\right)=kx\)
\(\Rightarrow\)\(51f\left(x_1\right)=51kx_1\) và \(2014f\left(x_2\right)=2014kx_2\)
\(\Rightarrow\)\(51f\left(x_1\right)-2014f\left(x_2\right)=51kx_1-2014kx_2\)\(=k\left(51x_1-2014x_2\right)=f\left(51x_1-2014x_2\right)\)