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a: ĐKXĐ: \(x\notin\left\{2;-2;0\right\}\)
b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{-x^2\left(x-2\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=x+1\\x\left(x+1\right)=-\left(x+1\right)\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\left(x+1\right)\left(x-1\right)=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2+42=4320\)
\(\Rightarrow x^2+90x-48x-4320=0\)
\(\Rightarrow x\left(x+90\right)-48\left(x-90\right)\)
\(\Rightarrow\left(x+90\right)\left(x-48\right)\)
\(\Rightarrow\orbr{\begin{cases}x+90=0\\x-48=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-90\\x=48\end{cases}}}\)
Ta có x^2+42x=4320
=>x^2+90x-48x-4320=0
=>x(x+90)-48(x+90)=0
=>(x-48)(x+90)=0
=>x=48 hoặc x=-90
Ta có:\(\left(x-3\right)\left(x+3\right)< x\left(x+2\right)+3\)
\(\Leftrightarrow x^2-9< x^2+2x+3\)
\(\Leftrightarrow x^2-x^2-2x< 3+9\)
\(\Leftrightarrow-2x< 12\)
\(\Leftrightarrow x>-6\)
Vậy tập nghiệm của BPT (1) là \(S=\left\{x\in R|x>-6\right\}\)
<=> (10x+8)/12-(2x-1)/12>48/12
<=>10x+8-2x+1>48
<=> 10x-2x>48-8-1
<=>8x>39
<=> x>39/8
Vậy tập n là {x/x>39/8}
\(\Leftrightarrow7\left(2x-1\right)-15x=-3x\)
=>14x-7-15x+3x=0
=>2x=7
hay x=7/2(nhận)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
\(\dfrac{7}{3x}-\dfrac{5}{2x-1}=\dfrac{1}{1-2x}\)
\(\Rightarrow\dfrac{7}{3x}+\dfrac{5}{1-2x}=\dfrac{1}{1-2x}\)
\(\Rightarrow\dfrac{7}{3x}=\dfrac{1}{1-2x}-\dfrac{5}{1-2x}\)
\(\Rightarrow\dfrac{7}{3x}=\dfrac{-4}{1-2x}\)
\(\Rightarrow-4.3x=7\left(1-2x\right)\)
\(\Rightarrow-12x=7-14x\)
\(\Rightarrow-12x+14x=7\)
\(\Rightarrow2x=7\)
\(\Rightarrow x=\dfrac{7}{2}\left(tm\right)\)