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\(a.\Leftrightarrow\frac{5x^2+16}{\left(x+4\right)\left(x-4\right)}=\frac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}DKXD:x\ne4;-4\)
\(\Rightarrow5x^2+16=2x^2-8x-x+4+3x^2+12x-x-4\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
\(b.\Leftrightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}.DKXD:y\ne2;-2\)
\(\Rightarrow y^2+2y+y+2-5y+10=12+y^2-4\)
\(\Leftrightarrow-2y=-4\)
\(\Leftrightarrow y=2\)
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
ĐK: \(x\ne-2;-3;-4;-5\)
\(1+\dfrac{1}{x+2}-\left(1+\dfrac{1}{x+3}\right)=1+\dfrac{1}{x+4}-\left(1+\dfrac{1}{x+5}\right)\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=\left(x+4\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+5x+6=x^2+9x+20\)
\(\Leftrightarrow4x=-14\Rightarrow x=-\dfrac{7}{2}\)
b/ ĐK: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow x^2+3x+2-\left(x^2+5x-14\right)-12=0\)
\(\Leftrightarrow-2x+4=0\Rightarrow x=2\) (ko t/m)
Vậy pt vô nghiệm
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
a) \(\left(2x-1\right)^2-\left(3x+5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-3x-5\right)=0\\ \text{}\Leftrightarrow\left(2x-1\right)\left(-x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-6\right\}\)
b) \(\dfrac{x+5}{4}-\dfrac{2x-3}{3}=\dfrac{2x-1}{12}\)
\(\Leftrightarrow3\left(x+5\right)-4\left(2x-3\right)=2x-1\\ \Leftrightarrow3x+15-8x+12=2x-1\\ \Leftrightarrow-5x+27=2x-1\\ \Leftrightarrow-5x-2x=-1-27\\ \Leftrightarrow-7x=-28\\ \Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
\(c)\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{-4}{1-x^2}\)(ĐKXĐ: \(x\ne\pm1\))
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \dfrac{\left(x+1\right)^2-\left(x-1\right)^2-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4x-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4}{x+1}=0\)
\(\Leftrightarrow4=0\)(vô lý)
Vậy .....
\(d)\dfrac{1}{x+1}+\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}\)(ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2}{x^2-x+1}=0\\ \Leftrightarrow\dfrac{x^2-x+1+2x-1-2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x+2x-2x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy ....
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x+2\right)\left(x-2\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow x^2+x+2x+2-5x+10=12+x^2-4\\ \Leftrightarrow-2x=-4\\ \Leftrightarrow x=2\left(ktm\right)\)
Vậy \(S\in\left\{\varnothing\right\}\)
ĐKXĐ: \(\begin{cases}x-2\ne 0\\x+2\ne 0\end{cases}\leftrightarrow x\ne 2\\x\ne -2\end{cases}\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\leftrightarrow \dfrac{(x+1)(x+2)}{(x-2)(x+2)}-\dfrac{5(x-2)}{(x+2)(x-2)}=\dfrac{12}{(x-2)(x+2)}+\dfrac{(x-2)(x+2)}{(x-2)(x+2)}\)
\(\to x^2+3x+2-5x+10=12+x^2-4\)
\(\leftrightarrow x^2-2x-x^2=12-12-4\)
\(\leftrightarrow -2x=-4\)
\(\leftrightarrow x=2(\rm KTM)\)
Vậy pt đã cho vô nghiệm \(S=\varnothing\)