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\(f\left(x_1\right)=ax_1\) ; \(f\left(x_2\right)=ax_2\) ; \(f\left(x_1x_2\right)=ax_1x_2\)
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 \(a=1\)
Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=a\cdot2^2+2b+c=4a+2b+c\\f\left(-5\right)=a\cdot\left(-5\right)^2-5b+c=25a-5b+c\end{matrix}\right.\)
\(\Rightarrow f\left(2\right)\cdot f\left(-5\right)=\left(4a+2b+c\right)\left(25a-5b+c\right)\)
Lại có:\(25a-5b+c=29a+2c-c-4a-5b\)
\(=3b-c-4a-5b=-2b-c-4a=-\left(4a+2b+c\right)\)
\(\Rightarrow f\left(2\right)\cdot f\left(-5\right)=-\left(4a+2b+c\right)\left(4a+2b+c\right)\)
\(=-\left(4a+2b+c\right)^2\le0\forall a,b,c\)
Nguyễn Việt Lâm Trần Trung Nguyên tran nguyen bao quan Shurima Azir Nguyễn Thanh Hằng Mysterious Person Phùng Khánh Linh Aki Tsuki
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) \(f\left(x_1-x_2\right)=f\left(x_1+\left(-x_2\right)\right)=f\left(x_1\right)+f\left(-x_2\right)=f\left(x_1\right)-f\left(x_2\right)\)
tham khảo
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0,3 x y + y = 6,5