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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\) <=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\) <=> \(25x+10-80x+10=24x+12-30\) <=> \(25x-80x-24x=12-30-10-10\) <=> \(-79x=-38\) <=> \(x=\dfrac{-38}{-79}\) \(x=\dfrac{38}{79}\) b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\) <=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\) <=> \(30x-12x+30+5x+40=210+10x-10\) <=> \(30x-12x+5x-10x=210-10-30-40\) <=> \(13x=130\) <=> \(x=\dfrac{130}{13}\) \(x=10\) c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\) <=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\) <=> \(28x+28+60x+120+105x+420+2520=0\) <=> \(28x+60x+105x=-28-120-420-2520\) <=> \(193x=-3088\) <=> \(x=\dfrac{-3088}{193}\) \(x=-16\) d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\) <=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\) <=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\) <=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\) <=> \(22968x=8199576\) <=> \(x=\dfrac{8199576}{22968}\) \(x=357\) \(a,\frac{-x}{4}+6=8\)\(\Leftrightarrow\frac{x}{-4}=2\Leftrightarrow x=-8\) b,\(\frac{-4}{x}-7=-5\Leftrightarrow\frac{-4}{x}=2\Leftrightarrow x=-2\) c,\(12+\frac{-6}{5x}=17\Leftrightarrow-\frac{6}{5x}=5\Leftrightarrow x=-\frac{6}{25}\) d,\(\frac{3-x}{7}=\frac{x+5}{4}\Leftrightarrow12-4x=7x+35\Leftrightarrow-11x=23\Leftrightarrow x=-\frac{23}{11}\) e,\(7-2x=-\frac{3}{3x}=-\frac{1}{x}\Leftrightarrow7x-2x^2+1=0\) \(\Leftrightarrow-2\left(x^2+\frac{7}{2}x+\frac{49}{16}\right)+\frac{57}{8}=0\Leftrightarrow\left(x+\frac{7}{4}\right)^2=\frac{57}{16}\) \(\Rightarrow\left[{}\begin{matrix}x+\frac{7}{4}=\frac{\sqrt{57}}{16}\\x+\frac{7}{4}=-\frac{\sqrt{57}}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{57}-28}{16}\\x=\frac{-\sqrt{57}-28}{16}\end{matrix}\right.\) a, \(-\frac{x}{4}+6=8\) => \(-\frac{x}{4}=8-6=2\) => \(x=2.-4=-8\) Vậy \(x\in\left\{-8\right\}\) \(b,\frac{4}{-x}-7=-5\) => \(\frac{4}{-x}=-5+\left(-7\right)=-12\) => \(x=4:12=\frac{1}{3}\) Vậy \(x\in\left\{\frac{1}{3}\right\}\) \(c,12+\frac{-6}{5x}=17\) => \(-\frac{6}{5x}=17-12=5\) => \(5x=-6:5=-\frac{6}{5}\) => \(x=-\frac{6}{5}:5=\frac{6}{25}\) Vậy \(x\in\left\{\frac{6}{25}\right\}\) \(d,\frac{3-x}{7}=\frac{x+5}{4}\) =>\(4\left(3-x\right)=7\left(x+5\right)\) => \(12-4x=7x+35\) => \(-4x-7x=35-12\) => \(-11x=23\) => \(x=23:\left(-11\right)=-\frac{23}{11}\) Vậy \(x\in\left\{-\frac{23}{11}\right\}\) e, \(7-2x=-\frac{3}{3x}\) => \(7-2x=-\frac{1}{x}\) => \(7=2x+\left(-\frac{1}{x}\right)\) => \(7=2x-\frac{1}{x}\) => \(7=\frac{2x^2}{x}-\frac{1}{x}\) => \(7=\frac{2x^2-1}{x}\) => :))
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CM
Cao Minh Tâm
18 tháng 3 2018
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