a. \(x\ne5\) là ĐKXĐ của biểu thức P

b. P =\(\dfrac{\left(x-5\right)^2}{x-5}\)=\(x-5\)

c. P = -1 <=> x-5 =-1 <=> x=4

\(\dfrac{x}{x-5}+\dfrac{4x}{x+5}+\dfrac{x\left(x-15\right)}{x^2-25}\)

  = \(\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{4x\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x\left(x-15\right)}{\left(x+5\right)\left(x-5\right)}\)

  = \(\dfrac{x^2+5x+4x^2-20x+x^2-15x}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x^2-30x}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)

  = \(\dfrac{6x}{x+5}\)

\(đk:x\ne1\\ \dfrac{x^2+5x}{3x^2-6x+3}:\dfrac{7x+35}{6x-6}\\ =\dfrac{x\left(x+5\right)}{3\left(x^2-2x+1\right)}:\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{x\left(x+5\right)}{3\left(x-1\right)^2}\times\dfrac{6\left(x-1\right)}{7\left(x+5\right)}\\ =\dfrac{2x}{7\left(x-1\right)}\)

\(đk:x\ne1\)

\(\dfrac{x^2+5}{3x^2-6x+3}.\dfrac{7x+35}{6x-6}\\ =\dfrac{x^2+5}{3\left(x^2-2x+1\right)}.\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{x^2+5}{3\left(x-1\right)^2}.\dfrac{7\left(x+5\right)}{6\left(x-1\right)}\\ =\dfrac{7\left(x^2+5\right)\left(x+5\right)}{18.\left(x-1\right)^3}\)

\(đk:x\ne0;1;-1\\ \dfrac{x^2+x}{x^2-x}.\dfrac{2x^2-2x}{x+1}\\ =\dfrac{x\left(x+1\right)}{x\left(x-1\right)}.\dfrac{2x\left(x-1\right)}{x+1}\\ =2x\)

\(đk:x\ne1\\ \dfrac{3x}{x-1}-\dfrac{5x+1}{2x-2}\\ =\dfrac{3x}{x-1}-\dfrac{5x+1}{2\left(x-1\right)}\\ =\dfrac{2.3x}{2\left(x-1\right)}-\dfrac{5x+1}{2\left(x-1\right)}\\ =\dfrac{6x-5x-1}{2\left(x-1\right)}\\ =\dfrac{x-1}{2\left(x-1\right)}\\ =\dfrac{1}{2}\)

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Áp dụng định lý Bơ-du ta có:

f(x) ⋮ (2x+5) ⇔ f(\(\dfrac{-5}{2}\))=0 (Kiến thức nâng cao lớp 8)

                    ⇔ (\(\dfrac{-5}{2}\))⁴+(\(\dfrac{-5}{2}\))³+12.(\(\dfrac{-5}{2}\))²-m=0

                    ⇔ \(\dfrac{1575}{16}\)-m=0

                    ⇔ m= \(\dfrac{1575}{16}\)

Vậy m=\(\dfrac{1575}{16}\) thì x⁴+x³+12x²-m chia hết cho 2x+5

\(B=\dfrac{x^2-x+1+4}{x^2-x+1}=1+\dfrac{4}{x^2-x+1}=1+\dfrac{4}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Do \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4};\forall x\)

\(\Rightarrow B\le1+\dfrac{4}{\dfrac{3}{4}}=\dfrac{19}{3}\)

Vậy \(B_{max}=\dfrac{19}{3}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)