Question

[Nguyễn]

.Phiền các bạn ><

Answer 1

[Nguyễn Huy Tú]

a, \(P=\left(1+\frac{1}{x-1}\right)\left(\frac{x^2-7}{x^2-4x+3}+\frac{1}{x-1}+\frac{1}{3-x}\right)\)ĐK : \(x\ne1;3\)

\(=\left(\frac{x}{x-1}\right)\left(\frac{x^2-7+x-3-x+1}{\left(x-1\right)\left(x-3\right)}\right)=\frac{x}{x-1}.\frac{x^2-9}{\left(x-1\right)\left(x-3\right)}=\frac{x\left(x+3\right)}{\left(x-1\right)^2}\)

b, Ta có : \(\left|x+2\right|=5\)

TH1 : \(x+2=5\Leftrightarrow x=3\)( ktmđk )

TH2 : \(x+2=-5\Leftrightarrow x=-7\)( tmđk )

Thay x = -7 vào biểu thức P ta được : \(P=\frac{-7\left(-7+3\right)}{\left(-7-1\right)^2}=\frac{49-21}{64}=\frac{28}{64}=\frac{7}{16}\)

c, Ta có : \(P>1\Rightarrow\frac{x\left(x+3\right)}{\left(x-1\right)^2}>1\)

\(\Leftrightarrow\frac{x\left(x+3\right)}{\left(x-1\right)^2}-1>0\Leftrightarrow\frac{x^2+3x-x^2+2x-1}{\left(x-1\right)^2}>0\)

\(\Leftrightarrow\frac{5x-1}{\left(x-1\right)^2}>0\Rightarrow5x-1>0\Leftrightarrow x>\frac{1}{5}\)

Answer 2

[Ngô Chi Lan]

\(P=\left(1+\frac{1}{x-1}\right)\left(\frac{x^2-7}{x^2-4x+3}+\frac{1}{x-1}+\frac{1}{3-x}\right)\)

\(=\left(\frac{x-1+1}{x-1}\right)\left(\frac{x^2-7}{\left(x-3\right)\left(x-1\right)}+\frac{x-3}{\left(x-1\right)\left(x-3\right)}-\frac{x-1}{\left(x-1\right)\left(x-\right)}\right)\)

\(=\frac{x}{x-1}.\frac{x^2-7+x-3-x+1}{\left(x-3\right)\left(x-1\right)}\)

\(=\frac{x}{x-1}.\frac{x^2+5}{\left(x-1\right)\left(x-1\right)}\)

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