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\(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\Leftrightarrow\dfrac{x^2-2x-4}{x^2-2x-3}-1>0\)
\(\Leftrightarrow\dfrac{x^2-2x-4-x^2+2x+3}{x^2-3x+x-3}>0\)
\(\Leftrightarrow\dfrac{-1}{\left(x-3\right)\left(x+1\right)}>0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x>-1\end{matrix}\right.\end{matrix}\right.\)
TH1 : vô lý
Vậy \(-1< x< 3\) thì \(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\)
\(\Rightarrow\left(5x^2+3x-2\right)^2-\left(4x^2-3x-2\right)^2=0\)
\(\Rightarrow\left[\left(5x^2+3x-2\right)-\left(4x^2-3x-2\right)\right]\left[\left(5x^2+3x-2\right)+\left(4x^2-3x-2\right)\right]=0\)
\(\Rightarrow\left(5x^2+3x-2-4x^2+3x+2\right)\left(5x^2+3x-2+4x^2-3x-2\right)=0\)
\(\Rightarrow\left(x^2+6x\right)\left(9x^2-4\right)=0\)
\(\Rightarrow x\left(x+6\right)\left[\left(3x\right)^2-2^2\right]=0\)
\(\Rightarrow x\left(x+6\right)\left(3x-2\right)\left(3x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+6=0\\3x-2=0\\3x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\3x=2\\3x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2.a)\(\dfrac{3\text{x}-2}{2}\)=\(\dfrac{1-2\text{x}}{3}\)
<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)
<=>9x-6=2-4x
<=>9x+4x=2+6
<=>13x=8
<=>x=\(\dfrac{8}{13}\)
1.a)2(x-0,5)+3=0,25(4x-1)
<=>2x-1+3=x-1phần4
<=>2x-x=-1/4+1-3
<=>x=-3/4
\(\dfrac{3x+2}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{x^2+2x+1}\)
= \(\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}-\dfrac{3x-2}{\left(x+1\right)^2}\)
= \(\dfrac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\dfrac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)^2}-\dfrac{\left(3x-2\right)\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\)
= \(\dfrac{3x^3+8x^2+7x+2}{\left(x^2-1\right)^2}-\dfrac{6x^2-6}{\left(x^2-1\right)^2}-\dfrac{3x^3-8x^2+7x-2}{\left(x^2-1\right)^2}\)
= \(\dfrac{10x^2+10}{\left(x^2-1\right)^2}\)
= \(\dfrac{10\left(x^2+1\right)}{\left(x^2-1\right)^2}\)
a)\(\dfrac{7x-1}{2}+2x=\dfrac{16-x}{3}\)
\(\dfrac{\left(7x-1\right).3}{2.3}+\dfrac{2x.6}{6}=\dfrac{\left(16-x\right)2}{3.2}\)
khử mẫu
=> (7x-1).3+12x=(16-x).2
=>21x-3+12x=-2x+32
=>21x-3+12x+2x-32=0
=>35x-35=0
b)\(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
ĐKXĐ: x khác +-2
\(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)
khử mẫu
(x+1).(x+2)+(x-1)(x-2)=2x2+4
=>x2+x+2+x+2+x2-2x-x+2=2x2+4
=>x2+x+2+x+2+x2-2x-x+2-2x2-4=0
=>(x2+x2-2x2)+(x+x-2x-x)+(2+2+2-4)=0
=>-x+2=0
=>-x=-2
=>x=2(loại)
vậy pt vô nghiệm
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{x^2+3}{x^2-2x}\)
<=> \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{x^2+3}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)-x+2}{x\left(x-2\right)}=\frac{x^2+3}{x\left(x-2\right)}\)
=> x2+2x-x+2=x2+3
<=>x=3
<=> \(\dfrac{x+2}{x-2}\)-\(\dfrac{1}{x}\)=\(\dfrac{2}{x\left(x-2\right)}\)
<=> \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
ok, ở đây đã có mẫu chung rồi, em cứ vậy làm tiếp thôi :D
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (ĐKXĐ: \(x\ne0;x\ne2\))
\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2-2=0\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow S=\left\{-1\right\}\)