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Lời giải:
\(\frac{x^2-4x+4}{4-x^2}=\frac{x^2-2.2.x+2^2}{2^2-x^2}=\frac{(x-2)^2}{(2-x)(2+x)}=\frac{(2-x)^2}{(2-x)(2+x)}=\frac{2-x}{2+x}\) (đpcm)
\(\frac{x^3-9x}{15-5x}=\frac{x(x^2-9)}{5(3-x)}=\frac{x(x-3)(x+3)}{5(3-x)}=\frac{-x(3-x)(x+3)}{5(3-x)}=\frac{-x(x+3)}{5}=\frac{-x^2-3x}{5}\) (đpcm)
a: ĐKXĐ: x<>2; x<>-2
b: \(A=\dfrac{3x\left(x-2\right)+2x+6}{2\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2-6x+2x+6}{2\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x^2+4x+6}{2\left(x-2\right)\left(x+2\right)}\)
c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)
Bài 1:
\(\dfrac{x^3-9x}{15-5x}=-\dfrac{x\left(x^2-9\right)}{5\left(x-3\right)}=\dfrac{-x\left(x-3\right)\left(x+3\right)}{5\left(x-3\right)}=\dfrac{-x\left(x+3\right)}{5}=\dfrac{-x^2-3x}{5}\)
Bài 2:
Sửa đề: \(\dfrac{4x^2-3x-7}{A}=\dfrac{4x-7}{2x+3}\)
\(\Leftrightarrow A=\dfrac{\left(4x^2-3x-7\right)\left(2x+3\right)}{4x-7}\)
\(=\dfrac{4x^2-7x+4x-7}{4x-7}\cdot\left(2x+3\right)\)
\(=\left(x+1\right)\left(2x+3\right)\)
\(\dfrac{3}{x-2}=\dfrac{3x}{\left(x-2\right)x}=\dfrac{3x}{x^2-2x}\)
Ta có biến đổi sau :
\(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x^2+x-3x-3}{x\left(x+1\right)}=\dfrac{x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\left(1\right)\)Tương tự , ta có :
\(\dfrac{x^2-4x+3}{x^2-x}=\dfrac{x^2-x-3x+3}{x\left(x-1\right)}=\dfrac{x\left(x-1\right)-3\left(x-1\right)}{x\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\left(2\right)\)Do đó , ba phân thức bằng nhau