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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(\left(x^2-4\right)+\left(8-5.x\right).\left(x+2\right)+4.\left(x-2\right).\left(x+1\right)=0\) \(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+\left(4.x-8\right).\left(x+1\right)=0\) \(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+4.x^2+4.x-8.x-8=0\) \(\Leftrightarrow0+4-6.x=0\) \(\Leftrightarrow4-6.x=0\) \(\Leftrightarrow-6.x=-4\) \(\Rightarrow x=\frac{2}{3}\) Vậy x = \(\frac{2}{3}\) a) 7x - 35 = 0 <=> 7x = 0 + 35 <=> 7x = 35 <=> x = 5 b) 4x - x - 18 = 0 <=> 3x - 18 = 0 <=> 3x = 0 + 18 <=> 3x = 18 <=> x = 5 c) x - 6 = 8 - x <=> x - 6 + x = 8 <=> 2x - 6 = 8 <=> 2x = 8 + 6 <=> 2x = 14 <=> x = 7 d) 48 - 5x = 39 - 2x <=> 48 - 5x + 2x = 39 <=> 48 - 3x = 39 <=> -3x = 39 - 48 <=> -3x = -9 <=> x = 3 \(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\) \(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\) \(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\) \(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\) \(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\) \(=\frac{2x+1}{x-3}\) 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 \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\) 1) (2x - 3)2 = 4x2 - 8 <=> 4x2 - 12x + 9 = 4x2 - 8 <=> 12x + 9 = -8 <=> 12x = -17 <=> x = 17/12 1) (2x - 3)^2 = 4x^2 - 8 <=> 4x^2 - 12x + 9 = 4x^2 - 8 <=> 4x^2 - 12x + 9 - 4x^2 = -8 <=> -12x + 9 = -8 <=> -12x = -8 - 9 <=> -12x = -17 <=> x = 17/12 2) x - (x + 2)(x - 3) = 4 - x^2 <=> x - x^2 + 3x - 2x + 6 = 4 - x^2 <=> 2x - x^2 + 6 = 4 - x^2 <=> 2x - x^2 + 6 + x^2 = 4 <=> 2x + 6 = 4 <=> 2x = 4 + 6 <=> 2x = 10 <=> x = 5 3) 3x - (x - 3)(x + 1) = 6x - x^2 <=> 3x - x^2 - x + 3x + 3 = 6x - x^2 <=> 5x - x^2 + 3 = 6x - x^2 <=> 5x - x^2 + 3 + x^2 = 6x <=> 5x + 3 = 6x <=> 3 = 6x - 5x <=> 3 = x 4) 3x/4 = 6 <=> 3x = 6.4 <=> 3x = 24 <=> x = 8 5) 7 + 5x/3 = x - 2 <=> 21 + 5x = 3x - 6 <=> 5x = 3x - 6 - 21 <=> 5x = 3x - 27 <=> 5x - 3x = -27 <=> 2x = -27 <=> x = -27/2 6) x + 4 = 2/5x - 3 <=> 5x + 20 = 2x - 15 <=> 5x + 20 - 2x = -15 <=> 3x + 20 = -15 <=> 3x = -15 - 20 <=> 3x = -35 <=> x = -35/3 7) 1 + x/9 = 4/3 <=> x/9 = 4/3 - 1 <=> x/9 = 1/3 <=> x = 3
K
Khách
CM
Cao Minh Tâm
15 tháng 1 2019
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