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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 \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\) 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 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 a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2 \(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\) \(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\) Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\) Ta được : \(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\) Từ đó ta lắp vào ( 1 ) tính được x 1) a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\) (đk:x khác \(\frac{1}{2}\)) \(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\) Vậy x=\(\frac{25}{7}\) b) /7-2x/=x-3 \(x\ge\frac{7}{2}\) (đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\)) \(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\) Vậy x=4 2) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\) \(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\) \(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\) a) 2x^2 + 3 = 2x(x + 4) - 7 <=> 2x^2 + 3 = 2x^2 + 8x - 7 <=> 2x^2 - 2x^2 - 8x = - 7 - 3 <=> -8x = -10 <=> x = -10/-8 = 5/4 b) 4x^2 - 12x + 5 = 0 <=> 4x^2 - 2x - 10x + 5 = 0 <=> 2x(2x - 1) - 5(2x - 1) = 0 <=> (2x - 5)(2x - 1) = 0 <=> 2x - 5 = 0 hoặc 2x - 1 = 0 <=> x = 5/2 hoặc x = 1/2 c) |5 - 2x| = 1 - x +) nếu x >= 5/2, ta có: 5 - 2x = 1 - x <=> -2x + 1 = 1 - 5 <=> -x = -4 <=> x = 4 (tm) +) nếu x < 5/2, ta có: -(5 - 2x) = 1 - x <=> -5 + 2x = 1 - x <=> 2x + 1 = 1 + 5 <=> 3x = 6 <=> x = 2 (ktm) d) \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}-\frac{2x+3}{x^2+x+1}\) ; ĐKXĐ: x # 1 <=> \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x+3}{x^2+x+1}\) <=> \(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\) <=> 2(x^2 + x + 1) = (2x - 1)(2x + 1) - (2x + 3)(x - 1) <=> 2x^2 + 2x + 2 = 2x^2 - x + 2 <=> 2x^2 - 2x^2 + 2x - x = 2 - 2 <=> x = 0 mạn phép vô đây để kiếm câu trả lời \(2x^2+3=2x\left(x+4\right)-7\) \(< =>2x^2+3=2x.x+4.2x-7\) \(< =>2x^2+3=2x^2+8x-7\) \(< =>2x^2+3-2x^2=8x-7\) \(< =>\left(2x^2-2x^2\right)-8x=-7-3\) \(< =>-8x=-10< =>8x=10\) \(< =>x=10:8=\frac{10}{8}=\frac{5}{4}\)
K
Khách
CM
Cao Minh Tâm
20 tháng 11 2018
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Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
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<=> \(\hept{\begin{cases}5-2x\text{ nếu }5-2x\ge0\Leftrightarrow x\ge\frac{5}{2}\\-\left(5-2x\right)\text{ nếu }5-2x< 0\Leftrightarrow x< \frac{5}{2}\end{cases}}\)
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