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Trước hết xoá \(\frac{2x}{a^2-a+1}\)ở 2 vế. Nếu \(\frac{a}{a+1}>0\left(a< -1;a>0\right)\)thì \(x< \frac{a}{4}\). Nếu \(\frac{a}{a+1}< 0\left(-1< a< 0\right)\)thì \(x>\frac{a}{4}\)
\(ĐKXĐ:a\ne-1\)
\(\frac{2x}{a^2-a+1}-\frac{1}{2a+2}< \frac{4x-1}{2a^2-2a+2}+\frac{a-2ax}{1+a^3}\Leftrightarrow\frac{2x}{a^2-a+1}-\frac{1}{2a+2}< \frac{2x}{a^2-a+1}-\frac{1}{2a^2-2a+2}+\frac{a}{1+a^3}-\frac{2ax}{1+a^3}\)\(\Leftrightarrow\frac{1}{2a+2}-\frac{1}{2a^2-2a+2}+\frac{a}{1+a^3}>\frac{2ax}{1+a^3}\Leftrightarrow\frac{a^2-a+1-a-1+2a}{2\left(a^3+1\right)}>\frac{2ax}{1+a^3}\Leftrightarrow\frac{a^2}{2\left(1+a^3\right)}>\frac{4ax}{2\left(1+a^3\right)}\)\(\Leftrightarrow\frac{4ax}{a+1}< \frac{a^2}{a+1}\)
* Nếu \(\frac{a}{a+1}>0\)(tức là a < -1 hoặc a > 0) thì \(x< \frac{a}{4}\)
* Nếu \(\frac{a}{a+1}< 0\)(tức là -1 < a < 0) thì \(x>\frac{a}{4}\)
\(\Leftrightarrow\dfrac{2x}{a^2-a+1}+\dfrac{-4x}{2a^2-2a+2a^2}+\dfrac{2ax}{1+a^3}< \dfrac{1}{2a+2}-\dfrac{1}{2a^2-2a+2}+\dfrac{a}{1+a^3}\)
\(\Leftrightarrow\left(\dfrac{2}{a^2-a+1}-\dfrac{4}{2a^2-2a+2}+\dfrac{2a}{1+a^3}\right).x< \left(\dfrac{1}{2a+2}-\dfrac{1}{2a^2-2a+2}+\dfrac{a}{1+a^3}\right)\)
\(\Leftrightarrow\left(\dfrac{2a}{1+a^3}\right)x< \dfrac{\left(a^2-a+1\right)-\left(a+1\right)+2a}{2.\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{a^2}{1+a^3}\)
\(\Leftrightarrow\left(\dfrac{2a}{1+a^3}\right)x< \dfrac{a^2}{2.\left(1+a^3\right)}\)
\(a=0\Rightarrow vo...N_o\)
\(\left\{{}\begin{matrix}\dfrac{2a}{a^3+1}>0\Leftrightarrow\left[{}\begin{matrix}a< -1\\a>0\end{matrix}\right.\\x< \dfrac{a^2}{2\left(a^3+1\right)}:\dfrac{2a}{\left(a^3+1\right)}=\dfrac{a}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2a}{a^3+1}< 0\Rightarrow-1< a< 0\\x>\dfrac{a}{2}\end{matrix}\right.\)
a, Ta có: \(a\left(ax-1\right)=x-1\)
\(\Leftrightarrow a^2x-a=x-1\)
\(\Leftrightarrow a^2x-x=a-1\)
\(\Leftrightarrow x\left(a-1\right)\left(a+1\right)=a-1\)
Với \(a\ne\pm1\)=> Pt có nghiệm duy nhất \(x=\frac{a-1}{a+1}\)
Với \(a=1\)=> Pt có nghiệm đúng với mọi x
Với \(a=-1\)=> Pt vô nghiệm
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
