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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\) \(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\) \(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\) \(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\) \(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\) \(=\frac{2x+1}{x-3}\) 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 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 \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\) \(\left(x^2-4\right)+\left(8-5.x\right).\left(x+2\right)+4.\left(x-2\right).\left(x+1\right)=0\) \(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+\left(4.x-8\right).\left(x+1\right)=0\) \(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+4.x^2+4.x-8.x-8=0\) \(\Leftrightarrow0+4-6.x=0\) \(\Leftrightarrow4-6.x=0\) \(\Leftrightarrow-6.x=-4\) \(\Rightarrow x=\frac{2}{3}\) Vậy x = \(\frac{2}{3}\) a, \(C=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\) \(C=5.\left(4x^2-4x+1\right)+4\left(x^2+3x-x-3\right)-2.\left(25-75x+9x^2\right)\) \(C=20x^2-20x+5+4x^2+8x-12-50+150x-18x^2\) \(=\left(20x^2+4x^2-18x^2\right)+\left(-20x+8x+150x\right)+\left(5-12-50\right)\) \(C=6x^2+138x-57\) Chúc bạn học tốt!!! Cũng không chắc có đúng hay sai nữa do cồng kềnh quá ! \(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\) \(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\) Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\) a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\) \(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\) \(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\) \(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0 \(x-1=0\) \(x=1\) \(a,ĐKXĐ:x\ne\pm\frac{1}{2}\) Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\) \(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\) \(\Leftrightarrow4x-2-6x-3=4\) \(\Leftrightarrow-2x=9\) \(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ) Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\) \(b,ĐKXĐ:x\ne\pm1;-3\) Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\) \(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\) \(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\) \(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\) \(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\) \(\Leftrightarrow9x^2+14x+13=0\) \(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\) \(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\) Pt vô nghiệm \(c,ĐKXĐ:x\ne1\) Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\) \(\Leftrightarrow x^2+x+1+2x^2-5=x-1\) \(\Leftrightarrow3x^2=3\) \(\Leftrightarrow x^2=1\) \(\Leftrightarrow x=\pm1\) Kết hợp vs ĐKXĐ được x = -1 Vậy pt có nghiệm duy nhất x = -1 làm lần lượt nha(bài nào k bt bỏ qua) \(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\) \(\Rightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\) \(\Rightarrow-2x-5=4\) \(\Rightarrow-2x=9\) \(\Rightarrow x=\frac{9}{-2}\)
K
Khách
CM
Cao Minh Tâm
30 tháng 8 2017
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