dùng định nghĩa 2 phân tử bằng nhau tìm đa thức A
\(\frac{5x^2-13x+6}{A}\) = \(\frac{5x-3}{2x+5}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{5x^2-13x+6}{A}=\frac{5x-3}{2x+5}\)
\(\Rightarrow\left(5x^2-13x+6\right)\left(2x+5\right)=A.\left(5x-3\right)\)
\(\Rightarrow\left[5x\left(x-2\right)-3\left(x-2\right)\right]\left(2x+5\right)=A.\left(5x-3\right)\)
\(\Rightarrow\left(x-2\right)\left(5x-3\right)\left(2x+5\right)=A.\left(5x-3\right)\)
\(\Rightarrow A=\left(x-2\right)\left(2x+5\right)=2x^2+x-10\)
a)\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=\frac{A}{\left(x-1\right)^2}\)
\(\Leftrightarrow\frac{x+4}{x-1}=\frac{A}{\left(x-1\right)^2}\). Nhân 2 vế ở tử với x-1 ta có:
\(x+4=\frac{A}{x-1}\Leftrightarrow A=\left(x-1\right)\left(x+4\right)=x^2+3x-4\)
b)\(\frac{x^2-3x}{2x^2-7x+3}=\frac{x^2+4x}{A}\)
\(\Leftrightarrow\frac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}=\frac{x\left(x+4\right)}{A}\)
\(\Leftrightarrow\frac{x}{2x-1}=\frac{x\left(x+4\right)}{A}\).Nhân 2 vế ở mẫu với x ta có:
\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)
\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)
\(\dfrac{A}{x-3}=\dfrac{y-x}{3-x}\)
\(\Rightarrow A=\dfrac{\left(x-3\right)\left(y-x\right)}{3-x}\)
\(\Rightarrow A=\dfrac{-\left(3-x\right)\left(y-x\right)}{3-x}\)
\(\Rightarrow A=x-y\)
_____
\(\dfrac{5x}{x+1}=\dfrac{Ax\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\)
\(\Rightarrow A=\dfrac{5x\left(x+1\right)\left(1-x\right)}{x\left(x+1\right)}\)
\(\Rightarrow A=5\left(1-x\right)\)
\(\Rightarrow A=5-5x\)
____
\(\dfrac{4x^2-5x+1}{A}=\dfrac{4x-1}{x+3}\)
\(\Rightarrow\dfrac{\left(4x-1\right)\left(x-1\right)}{A}=\dfrac{4x-1}{x+3}\)
\(\Rightarrow A=\dfrac{\left(4x-1\right)\left(x-1\right)\left(x+3\right)}{4x-1}\)
\(\Rightarrow A=\left(x-1\right)\left(x+3\right)\)
\(\Rightarrow A=x^2+2x-3\)
Ta có:
\(5\left(x^3-9x\right)=5x^3-45x.\)(1)
\(\left(15-5x\right).\left(-x^2-3x\right)=-15x^2-45x+5x^3+15x^2=5x^3-45x\)(2)
Từ (1)(2) suy ra \(5\left(x^3-9x\right)=\left(15-5x\right)\left(-x-3x\right)\)
\(\Rightarrow\frac{x^3-9x}{15-5x}=\frac{-x^2-3x}{5}\)(Điều phải chứng minh)