Bài 1: Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}\) Tính GTBT: \(f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+...+f\left(\frac{2016}{2017}\right).\)Bài 2: Giải HPT sau: \(\hept{\begin{cases}2x^2-y^2+xy+3y=2\\x^2-y^2=3\end{cases}}\)Bài 3: Tìm m để PT: \(x^4+x^3+\left(m-2\right)x^2-4mx-2m^2=0\)có 4 nghiệm...
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Bài 1: Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}\)
Tính GTBT: \(f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+...+f\left(\frac{2016}{2017}\right).\)
Bài 2: Giải HPT sau: \(\hept{\begin{cases}2x^2-y^2+xy+3y=2\\x^2-y^2=3\end{cases}}\)
Bài 3: Tìm m để PT: \(x^4+x^3+\left(m-2\right)x^2-4mx-2m^2=0\)có 4 nghiệm thỏa \(x_1^2+x_2^2+x_3^2+x_4^2=5\)
Ta có: \(\frac{1}{f\left(x\right)}-1=\frac{\left(1-x\right)^3}{x^3}\)
Xét hai số a, b dương sao cho \(a+b=1\)
Ta có: \(\hept{\begin{cases}\frac{1}{f\left(a\right)}-1=\frac{\left(1-a\right)^3}{a^3}\\\frac{1}{f\left(b\right)}-1=\frac{\left(1-b\right)^3}{b^3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{1-f\left(a\right)}{f\left(a\right)}=\frac{\left(1-a\right)^3}{a^3}\\\frac{1-f\left(b\right)}{f\left(b\right)}=\frac{a^3}{\left(1-a\right)^3}\end{cases}}\)
\(\Rightarrow\frac{1-f\left(a\right)}{f\left(a\right)}.\frac{1-f\left(b\right)}{f\left(b\right)}=1\)
\(\Rightarrow f\left(a\right)+f\left(b\right)=1\)
Áp dụng vào bài toán ta được
\(f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+...+f\left(\frac{2016}{2017}\right)\)
\(=\left[f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)\right]+\left[f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)\right]+...+\left[f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)\right]\)
\(=1+1+...+1=1008\)
Câu 2/
\(\hept{\begin{cases}2x^2-y^2+xy+3y=2\left(1\right)\\x^2-y^2=3\left(2\right)\end{cases}}\)
Ta có:
\(\left(1\right)\Leftrightarrow\left(x+y-1\right)\left(2x-y+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=1-x\\y=2x+2\end{cases}}\)
Thế ngược lại (1) giải tiếp sẽ ra nghiệm.