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a) \(y=f\left(x\right)=-\frac{1}{2}x\)
\(f\left(-2\right)=-\frac{1}{2}.\left(-2\right)=1\)
\(f\left(3\right)=-\frac{1}{2}.3=-\frac{3}{2}\)
b)
Cho \(x=1\Rightarrow y=-\frac{1}{2}.1=-\frac{1}{2}\)
\(\Rightarrow A\left(1;-\frac{1}{2}\right)\)
Hình ko đẹp lắm mong cậu thông cảm
\(f\left(x\right)=\frac{x^2+2x+1-x^2}{x^2\left(x+1\right)^2}=\frac{\left(x+1\right)^2-x^2}{x^2\left(x+1\right)^2}=\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)
\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)
\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)
a) \(f\left(\frac{-1}{2}\right)\)
Thay x = -1/2 vào ta được: \(y=f\left(\frac{-1}{2}\right)=\left(\frac{-1}{2}\right)^2-5.\left(\frac{-1}{2}\right)+1=\frac{15}{4}\)
\(f\left(3\right)\)
Thay x = 3 vào ta được: \(y=f\left(3\right)=3^2-5.3+1=-5\)
b) Để f(x) = 1
Suy ra: \(x^2-5x+1=1\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Vậy khi x = 0 hoặc x = 5 thì f(x) = 1
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