Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\Leftrightarrow\frac{5x^2+16}{\left(x+4\right)\left(x-4\right)}=\frac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}DKXD:x\ne4;-4\)
\(\Rightarrow5x^2+16=2x^2-8x-x+4+3x^2+12x-x-4\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
\(b.\Leftrightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}.DKXD:y\ne2;-2\)
\(\Rightarrow y^2+2y+y+2-5y+10=12+y^2-4\)
\(\Leftrightarrow-2y=-4\)
\(\Leftrightarrow y=2\)
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 \(S=\left\{-1\right\}\)
a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>3x+5<10x-30
=>-7x<-35
hay x>5
b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)
=>14x-80>-11x
=>25x>80
hay x>16/5
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
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
\(a,3x^2+2x-1\Leftrightarrow3x^2-x+3x-1\Leftrightarrow x\left(3x-1\right)+\left(3x-1\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)\)
a)
\(\dfrac{5x^2+16}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\) (\(x\ne\pm2\))
\(\Rightarrow\dfrac{5x^2+16}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Rightarrow\dfrac{5x^2+16-\left(2x^2-8x-x+4\right)-\left(3x^2+12x-x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Rightarrow\dfrac{10x+16}{x^2-16}=0\)
=> 10x + 16 =0
=> 10x = -16
=> x = \(-\dfrac{8}{5}\)
1.
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\left(ĐKXĐ:x\ne1\right)\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\ \Leftrightarrow21x-9=2x-2\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\dfrac{7}{19}\left(TMĐK\right)\)
2.
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\left(ĐKXĐ:x\ne-\dfrac{2}{3};x\ne\dfrac{1}{3}\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\\ \Leftrightarrow-8x+1=-11x-14\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\left(TMĐK\right)\)
3.
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ \Leftrightarrow\left(\dfrac{1-x}{x+1}+3\right)\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}.\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{4+2x}{x+1}\left(x+1\right)=2x+3\\ \Leftrightarrow4+2x=2x+3\\ \Leftrightarrow4=3\)
Vô nghiệm.
\(\Leftrightarrow x-2+5x+10=3x-12\)
=>6x+8=3x-12
=>3x=-20
hay x=-20/3(nhận)