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6 tháng 3 2020
\(\left(x^2-4\right)+\left(8-5.x\right).\left(x+2\right)+4.\left(x-2\right).\left(x+1\right)=0\)
\(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+\left(4.x-8\right).\left(x+1\right)=0\)
\(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+4.x^2+4.x-8.x-8=0\)
\(\Leftrightarrow0+4-6.x=0\)
\(\Leftrightarrow4-6.x=0\)
\(\Leftrightarrow-6.x=-4\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 12 2023
a, 6\(x^2\) - (2\(x\) - 3).(3\(x\) + 2) = 1
6\(x^2\) - (6\(x^2\) + 4\(x\) - 9\(x\) - 6) = 1
6\(x^2\) - 6\(x^2\) - 4\(x\) + 9\(x\) + 6 = 1
(6\(x^2\) - 6\(x^2\)) + (9\(x\) - 4\(x\)) + 6 = 1
5\(x\) + 6 = 1
5\(x\) = 1 - 6
5\(x\) = -5
\(x\) = - 5 : 5
\(x\) = - 1
b, (\(x\) + \(\dfrac{1}{2}\))2 - (\(x\) + \(\dfrac{1}{2}\)).(\(x\) + 6) = 8
\(x^2\) + \(x\) + \(\dfrac{1}{4}\) - (\(x^2\) + 6\(x\) + \(\dfrac{1}{2}\)\(x\) + 3) = 8
\(x^2\) + \(x\) + \(\dfrac{1}{4}\) - \(x^2\) - 6\(x\) - \(\dfrac{1}{2}\)\(x\) - 3 = 8
(\(x^2\) - \(x^2\)) + (\(x\) - 6\(x\) - \(\dfrac{1}{2}\)\(x\)) - ( 3 - \(\dfrac{1}{4}\)) = 8
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{11}{4}\) = 8
\(\dfrac{11}{2}\)\(x\) = - 8 - \(\dfrac{11}{4}\)
\(\dfrac{11}{2}\)\(x\) = - \(\dfrac{43}{4}\)
\(x\) = \(\dfrac{-43}{4}\) : \(\dfrac{11}{2}\)
\(x\) = \(\dfrac{-43}{22}\)
9 tháng 3 2022
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
9 tháng 3 2022
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)