\(\left(x+a\right)\left(x+5\right)=x^2+3x+b\)

\(\Leftrightarrow x^2+5x+ax+5a=x^2+3x+b\)

\(\Leftrightarrow x^2+\left(5+a\right)x+5a=x^2+3x+b\)

Đồng nhất ta có : \(\hept{\begin{cases}5+a=3\\5a=b\end{cases}\Rightarrow\hept{\begin{cases}a=-2\\b=-10\end{cases}}}\)

sao vế trái không có b

Tại bài toán vậy mà

vào câu hỏi tương tự khác có

 Ta có:\(VT=\left(x+a\right)\left(x+5\right)\)

\(=x^2+\left(a+5\right)x+5a\)

Và \(VP=x^2+3x+b\)

Đồng nhất 2 vế ta có: \(\hept{\begin{cases}a+5=3\\b=5a\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-2\\b=-10\end{cases}}\)

\(\dfrac{3x+1}{x^2-3x+2}=\dfrac{a}{x-1}+\dfrac{b}{x-2}\)

\(\Leftrightarrow\dfrac{3x+1}{x-1}-\dfrac{3x+1}{x-2}=\dfrac{a}{x-1}+\dfrac{b}{x-2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=3x+1\\b=-3x-1\end{matrix}\right.\)

em cảm ơn nhìu ạ

Theo c) \(f\left(\frac{5}{7}\right)=f\left(\frac{2}{7}+\frac{3}{7}\right)=f\left(\frac{2}{7}\right)+f\left(\frac{3}{7}\right)\)

      \(f\left(\frac{2}{7}\right)=f\left(\frac{1}{7}+\frac{1}{7}\right)=f\left(\frac{1}{7}\right)+f\left(\frac{1}{7}\right)=2.f\left(\frac{1}{7}\right)\) 

  \(f\left(\frac{3}{7}\right)=f\left(\frac{1}{7}+\frac{2}{7}\right)=f\left(\frac{1}{7}\right)+f\left(\frac{2}{7}\right)=f\left(\frac{1}{7}\right)+2f\left(\frac{1}{7}\right)=3.f\left(\frac{1}{7}\right)\)

       \(\implies\)\(f\left(\frac{5}{7}\right)=5.f\left(\frac{1}{7}\right)\)          (1)

 Theo b) \(f\left(\frac{1}{7}\right)=\frac{1}{7^2}.f\left(7\right)\)          (2)

Theo c) \(f\left(7\right)=f\left(3+4\right)=f\left(3\right)+f\left(4\right)\)

                                                 \(=2.f\left(3\right)+f\left(1\right)\) 

                                                 \(=6.f\left(1\right)+f\left(1\right)\) 

                                                 \(=7.f\left(1\right)\)

Theo a)\(f\left(1\right)=1\)\(\implies\)\(f\left(7\right)=7\)      (3)

    Từ (1);(2);(3)

       \(\implies\)       \(f\left(\frac{5}{7}\right)=\frac{5}{7}\)

︵✰He❤lloღ

<=> x^2+(a+5)x+5a=x^2+3x+b

<=>

a+5=3; a=-2

b=5a=-10