Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\) \(\left(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne-3\end{matrix}\right.\right)\)
\(\Leftrightarrow\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{\left(x-1\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(3x^2+8x-3\right)-\left(2x^2+3x-5\right)+4=x^2+2x-3\)
\(\Leftrightarrow x^2+5x+6=x^2+2x-3\)
\(\Leftrightarrow9=-3x\)
\(\Leftrightarrow x=-3\left(ktmđk\right)\)
\(\Leftrightarrow Ptvn\)
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
\(=>\frac{1-x+x+1}{x+1}+2=\frac{1}{x+1}+2\)
\(=>\frac{2}{x+1}=\frac{1}{x+1}\)
\(=>2x+2=x+1\)
\(=>2x-x=1-2=-1\)
\(=>x=-1\)
vậy nghiệm của phương trình trên là {-1}
À quên ĐKXĐ của câu a là \(x\ne-1\)
Nên \(x\in\varnothing\)nhé :v
\(\Leftrightarrow\frac{5-2x}{3\left(3x-1\right)}+\frac{3\left(x^2-1\right)}{3\left(3x-1\right)}-\frac{\left(x+2\right)\left(1-3x\right)}{3\left(3x-1\right)}=0\)
\(\Rightarrow5-2x+3x^2-3-x+3x^2-2+6x=0\Leftrightarrow6x^2+3x=0\Leftrightarrow3x\left(2x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow-\frac{5x-2}{2\left(x-1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}=\frac{1-x-x^2-x+3}{1-x}\Leftrightarrow\frac{2x^2-8x+3}{2\left(x-1\right)}=\frac{x^2+2x-4}{x-1}\)
\(\Rightarrow2x^2-8x+3=2x^2+4x-8\)\(\Leftrightarrow-8x+3=4x-8\Leftrightarrow-12x=-12\Rightarrow x=1\)