by Dùng định nghĩa hai phân thức bằng nhau, hãy tìm đa thức A trong mỗi đẳng thức sau:
a) \(\frac{{{x}^{2}}-4x-5}{A}=\frac{5-x}{x}\)
b) \(\frac{5{{x}^{2}}-13x+6}{A}=\frac{5x-3}{2x+5}\)
Lời giải chi tiết:
Hướng dẫn giải chi tiết
\(\begin{align} & a)\,\,\frac{{{x}^{2}}-4x-5}{A}=\frac{5-x}{x} \\ & \Rightarrow A(5-x)=x({{x}^{2}}-4x-5) \\& \Rightarrow A=x({{x}^{2}}-4x-5):(5-x) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=x({{x}^{2}}+x-5x-5):(5-x) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=x\left( x\left( x+1 \right)-5\left( x+1 \right) \right):(5-x) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=x(x+1)(x-5):(5-x) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=-x(x+1)(5-x):(5-x) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=-x(x+1) \\\end{align}\)
Vậy \(A=-x\left( x+1 \right).\)
\(b)\,\,\frac{5{{x}^{2}}-13x+6}{A}=\frac{5x-3}{2x+5}\)
Cách 1:
\(\begin{align} & \Rightarrow A(5x-3)=(5{{x}^{2}}-13x+6)(2x+5) \\ & \Rightarrow A=(5{{x}^{2}}-13x+6)(2x+5):(5x-3) \\ & \,\,\,\,\,\,\,\,\,\,\,\,=(10{{x}^{3}}+25{{x}^{2}}-26{{x}^{2}}-65x+12x+30):(5x-3) \\ & \,\,\,\,\,\,\,\,\,\,\,\,=(10{{x}^{3}}-{{x}^{2}}-53x+30):(5x-3) \\\end{align}\)
\(\begin{align} & 10{{x}^{3}}-{{x}^{2}}-53x+30\,\,\,\,\,\,\,\,\,\,\,5x-3 \\ & 10{{x}^{3}}-6{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{x}^{2}}+x-10 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5{{x}^{2}}-53x+30\, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5{{x}^{2}}-3x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-50x+30 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-50x+30 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\\end{align}\)
Cách 2:
\(\begin{align} & A=\left( 5{{x}^{2}}-13x+6 \right)\left( 2x+5 \right):\left( 5x-3 \right) \\ & \,\,\,\,\,=\left( 5{{x}^{2}}-10x-3x+6 \right)\left( 2x+5 \right):\left( 5x-3 \right) \\ & \,\,\,\,\,=\left( 5x\left( x-2 \right)-3\left( x-2 \right) \right)\left( 2x+5 \right):\left( 5x-3 \right) \\ & \,\,\,\,\,=\left( 5x-3 \right)\left( x-2 \right):\left( 5x-3 \right) \\ & \,\,\,\,\,=\left( x-2 \right)\left( 2x+5 \right) \\ & \,\,\,\,\,=2{{x}^{2}}+5x-4x-10 \\ & \,\,\,\,\,=2{{x}^{2}}+x-10. \\\end{align}\)
Vậy \(A=2{{x}^{2}}+x-10\).
