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Bài 16:
a: \(\dfrac{A}{x-2}=\dfrac{x^2+3x+2}{x^2-4}\)
=>\(\dfrac{A}{x-2}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+2\right)\left(x-2\right)}\)
=>\(\dfrac{A}{x-2}=\dfrac{x+1}{x-2}\)
=>A=x+1
b: \(\dfrac{M}{x-1}=\dfrac{x^2+3x+2}{x+1}\)
=>\(\dfrac{M}{x-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{x+1}\)
=>\(\dfrac{M}{x-1}=x+2\)
=>\(M=\left(x+2\right)\left(x-1\right)=x^2+x-2\)
Bài 17:
\(x^2-2x+4\)
\(=x^2-2x+1+3\)
\(=\left(x-1\right)^2+3>=3\forall x\)
=>\(P=\dfrac{15}{x^2-2x+4}< =\dfrac{15}{3}=5\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1