Bài 1
a) \(x^3-4x+3x^2-12\)
\(=\left(x^3+3x^2\right)-\left(4x+12\right)\)
\(=x^2\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-4\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
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b)
\(A=\dfrac{8x}{x+1}+\dfrac{8}{x+1}\)
ĐKXĐ: \(x\ne-1\)
\(A\)\(=\dfrac{8x+8}{x+1}\)
\(=\dfrac{8\left(x+1\right)}{x+1}\)
\(=8\)
\(B=\dfrac{x-5}{x+5}+\dfrac{2}{x-5}+\dfrac{5-x^2}{x^2-25}\)
ĐKXĐ: \(x\ne-5;x\ne5\)
\(B=\dfrac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{2\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{5-x^2}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2-10x+25+2x+10+5-x^2}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-8x+40}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-8\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-8}{x+5}\)
Bài 3
\(\Delta AHC\) vuông tại H
\(\Rightarrow AC^2=AH^2+CH^2\) (Pythagore)
\(\Rightarrow AH^2=AC^2-CH^2=17^2-8^2=225\)
\(\Rightarrow AH=15\left(km\right)\)
\(\Delta BHC\) vuông tại H
\(\Rightarrow BC^2=BH^2+CH^2\left(Pythagore\right)\)
\(\Rightarrow BH^2=BC^2-CH^2=10^2-8^2=36\)
\(\Rightarrow BH=6\left(km\right)\)
\(\Rightarrow\) Khoảng cách AB giữa hai bến tàu là:
\(AB=AH-BH=15-6=9\left(km\right)\)
Bài 2
\(C=\dfrac{3x+7}{x-5}\)
a) ĐKXĐ: \(x-5\ne0\)
\(x\ne5\)
b) Tại \(x=-9\), ta có:
\(C=\dfrac{3.\left(-9\right)+7}{-9+5}=\dfrac{-20}{-4}=5\)
