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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) 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 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, x( x - 1) = x ( x + 2) <=> x2 - x = x2 + 2x <=> x2 - x - x2 - 2x = 0 <=> -3x = 0 <=> x = 0 b, tương tự câu a c,\(\Leftrightarrow\frac{3x-3}{4}=2-\frac{x-2}{8}\) \(\Leftrightarrow\frac{\left(3x-3\right)2}{8}=\frac{16}{8}-\frac{x-2}{8}\) \(\Leftrightarrow\frac{6x-6}{8}=\frac{16}{8}-\frac{x-2}{8}\) => 6x - 6 = 16 - x + 2 <=> 6x + x = 16 + 2 + 6 <=> 7x = 24 <=> x=\(\frac{24}{7}\) Các câu còn lại làm tương tự a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\) \(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\) \(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\) \(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\) \(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\) Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\) \(\Rightarrow x+10=0\Rightarrow x=-10\) Vậy x = -10 b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\) \(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\) \(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\) \(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\) \(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\) Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0 Vậy x = 2012 a, (x+1)/9 +1 + (x+2)/8 = (x+3)/7 + 1 + (x+4)/6 + 1 <=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6 <=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0 vì 1/9 +1/8 -1/7 - 1/6 khác 0 => x+10=0 => x=-10 a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10 b, x/x-3 - x/x-5 = x/x-4 - x/x-6 c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0 d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6 e, 2/2x+1 - 3/2x-1 = 4/4x^2-1 f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3 g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1 1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\) \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\) \(\Leftrightarrow35x-5+60x=96-6x\) \(\Leftrightarrow95x-5=96-6x\) \(\Leftrightarrow95x+6x=96+5\) \(\Leftrightarrow101x=101\) \(\Leftrightarrow x=1\) 2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) \(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\) \(\Leftrightarrow30x+9=36+24+32x\) \(\Leftrightarrow30x+9=32x+60\) \(\Leftrightarrow30x-32x=60-9\) \(\Leftrightarrow-2x=51\) \(\Leftrightarrow x=-\frac{51}{2}\) 3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\) \(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\) \(\Leftrightarrow8x-3-6x+4=4x-2+x+3\) \(\Leftrightarrow2x+1=5x+1\) \(\Leftrightarrow2x=5x\) \(\Leftrightarrow x=0\) 4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\) => \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\) => \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\) => \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\) => \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\) => 27 - 9x + 80 - 16x = 12 - 12x - 48 => 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0 => (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0 => 143 - 13x = 0 => 13x = 143 => x = 11 5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\) => \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\) => \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\) => \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\) => \(\frac{6x-18+7x-35-13x-4}{21}=0\) => 6x - 18 + 7x - 35 - 13x - 4 = 0 => (6x + 7x - 13x) + (-18 - 35 - 4) = 0 => -57 = 0(vô nghiệm) 6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\) => \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\) => \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\) => \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\) => \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\) => \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\) => \(12x+10-10x-3=12x+2\) => \(2x+10-3=12x+2\) => 2x + 10 - 3 - 12x - 2 = 0 => (2x - 12x) + (10 - 3 - 2) = 0 => -10x + 5 = 0 => -10x = -5 => x = 1/2 7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\) => \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\) => \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\) => \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\) => 6x - 3 - 5x + 10 - x - 7 = 0 => (6x - 5x - x) + (-3 + 10 - 7) = 0 => 0x + 0 = 0 => 0x = 0 => x tùy ý Bài 8 tự làm nhé a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\) =>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\) =>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\) =>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\) =>x=0 b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\) =>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\) =>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\) Đến đây thì bạn giải giống câu a
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