BÀI 1cho biểu thức
P= (2√x+xx√x−1−1√x−1):(1−√x+2x+√x+1)(2x+xxx−1−1x−1):(1−x+2x+x+1)
a) Rút gọn P
b) tính √PP khi x=5+2√3
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a) đk: x khác 1; \(\dfrac{3}{2}\)
\(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)
= \(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{5-3x}{1-x}\)
= \(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}.\dfrac{1-x}{-3x+5}=\dfrac{-1}{2x-3}\)
b) Có \(\left|3x-2\right|+1=5\)
<=> \(\left|3x-2\right|=4\)
<=> \(\left[{}\begin{matrix}3x-2=4< =>x=2\left(Tm\right)\\3x-2=-4< =>x=\dfrac{-2}{3}\left(Tm\right)\end{matrix}\right.\)
TH1: Thay x = 2 vào P, ta có:
P = \(\dfrac{-1}{2.2-3}=-1\)
TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:
P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)
c) Để P > 0
<=> \(\dfrac{-1}{2x-3}>0\)
<=> 2x - 3 <0
<=> x < \(\dfrac{3}{2}\) ( x khác 1)
d) P = \(\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)
<=> 2x - 3 = x2 - 6
<=> x2 - 2x - 3 = 0
<=> (x-3)(x+1) = 0
<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne-\frac{1}{2}\end{cases}}\)
a) \(A=\left(\frac{1}{x-1}+\frac{x}{x^3-1}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow A=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow A=\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow A=\frac{x+1}{x-1}\)
b) Thay \(x=\frac{1}{2}\)vào A, ta được :
\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3\)
a: \(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b: Khi x=9 thì P=9-3+1=7
c: P=3
=>x-căn x-2=0
=>(căn x-2)(căn x+1)=0
=>x=4