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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, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\) ⇔ 3(x - 3) = 90 - 5(1 - 2x) ⇔ 3x - 9 = 90 - 5 + 10x ⇔ 3x - 10x = 90 - 5 + 9 ⇔ -7x = 94 ⇔ x = \(\frac{-94}{7}\) S = { \(\frac{-94}{7}\) } b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\) ⇔ 2(3x - 2) - 60 = 9 - 6(x + 7) ⇔ 6x - 4 - 60 = 9 - 6x - 42 ⇔ 6x + 6x = 9 - 42 + 60 + 4 ⇔ 12x = 31 ⇔ x = \(\frac{31}{12}\) S = { \(\frac{31}{12}\) } c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7 ⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210 ⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210 ⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40 ⇔ 13x = 150 ⇔ x = \(\frac{150}{13}\) S = { \(\frac{150}{13}\) } d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\) ⇔ 21x - 120(x - 9) = 4(2x + 1,5) ⇔ 21x - 120x + 1080 = 8x + 6 ⇔ 21x - 120x - 8x = 6 - 1080 ⇔ -107x = -1074 ⇔ x = \(\frac{1074}{107}\) S = { \(\frac{1074}{107}\) } e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5 ⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840 ⇔ 140x -140+56 -294x+42= 96x+48 -840 ⇔ 140x -294x -96x = 48 -840 -42 -56+140 ⇔ -250x = -750 ⇔ x = 3 S = { 3 } f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\) ⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x ⇔ 4x+4+18x+9 = 4x+6x+6+7+12x ⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4 ⇔ 0x = 0 S = R Chúc bạn học tốt ! Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html Mình cảm ơn trước nhaa Bài 1: \(a.5^5-5^4+5^3\) \(=5^3.5^2-5^3.5+5^3.1\) \(=5^3\left(5^2-5+1\right)\) \(=5^3.21\) \(=5^3.3.7⋮7\) \(\) \(\) \(\) Bài 2: \(a.32< 2^n< 128\) \(\Rightarrow2^5< 2^n< 2^7\) \(\Rightarrow n=2\) \(b.9.27\le3^n\le243\) \(\Rightarrow3^2.3^3\le3^n\le3^5\) \(\Rightarrow3^5\le3^n\le3^5\) \(\Rightarrow n=5\) Giải: 1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\) \(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\) \(=\dfrac{-1}{12}-\dfrac{55}{24}\) \(=\dfrac{-19}{8}\) 2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\) \(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\) \(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\) \(=\dfrac{5}{12}\) 3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\) \(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\) \(=-\dfrac{67}{120}\) 4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\) \(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\) \(=-\dfrac{43}{30}\) 5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\) \(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\) \(=\dfrac{3}{20}\) 6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\) \(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\) \(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\) \(=8+\dfrac{5}{8}\) \(=\dfrac{69}{8}\) 7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\) \(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\) \(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\) \(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\) \(=-\dfrac{1}{4}\left(13+6+1\right)\) \(=-\dfrac{1}{4}.20=-5\) 8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\) \(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\) \(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\) \(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\) \(=-7\left(6+1\right)\) \(=-7.7=-49\) Vậy ... 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 Bài 1: a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\) \(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\) \(\Leftrightarrow-5=0\)(vl) Vậy: \(x\in\varnothing\) b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\) \(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\) hay x=1 Vậy: x=1 c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\) \(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\) \(\Leftrightarrow2x-72=0\) \(\Leftrightarrow2\left(x-36\right)=0\) mà 2>0 nên x-36=0 hay x=36 Vậy: x=36 d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\) \(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\) \(\Leftrightarrow120x+36=56-64x\) \(\Leftrightarrow120x+36-56+64x=0\) \(\Leftrightarrow184x-20=0\) \(\Leftrightarrow184x=20\) hay \(x=\frac{5}{46}\) Vậy: \(x=\frac{5}{46}\) e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\) \(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\) \(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\) \(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\) \(\Leftrightarrow-23x+29=0\) \(\Leftrightarrow-23x=-29\) hay \(x=\frac{29}{23}\) Vậy: \(x=\frac{29}{23}\) f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\) \(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\) \(\Leftrightarrow2x+8-10x-50-25=0\) \(\Leftrightarrow-8x-67=0\) \(\Leftrightarrow-8x=67\) hay \(x=\frac{-67}{8}\) Vậy: \(x=\frac{-67}{8}\) g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\) \(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\) \(\Leftrightarrow10-5x-8x-8+12x-30=0\) \(\Leftrightarrow-x-28=0\) \(\Leftrightarrow-x=28\) hay x=-28 Vậy: x=-28 h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\) \(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\) \(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\) \(\Leftrightarrow0x=0\) Vậy: \(x\in R\) Bài 2: a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\) \(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\) \(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\) \(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\) \(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\) Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\) b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1) Ta có: \(x^2\ge0\forall x\) \(\Rightarrow x^2+5\ge5\ne0\forall x\)(2) Từ (1) và (2) suy ra: \(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\) Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\) c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\) \(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\) \(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\) \(\Leftrightarrow-x-3=0\) \(\Leftrightarrow-x=3\) hay x=-3 Vậy: Tập nghiệm S={-3} d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\) \(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\) \(\Leftrightarrow x^2-x-x^2-x+12-5x=0\) \(\Leftrightarrow12-7x=0\) \(\Leftrightarrow7x=12\) hay \(x=\frac{12}{7}\) Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\) e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x \(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\) \(\Leftrightarrow31x-1=0\) \(\Leftrightarrow31x=1\) hay \(x=\frac{1}{31}\) Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\) \(
a)\dfrac{{x - 3}}{5} = 6 - \dfrac{{1 - 2x}}{2}\\
\Leftrightarrow 2\left( {x - 3} \right) = 60 - 5\left( {1 - 2x} \right)\\
\Leftrightarrow 2x - 6 = 60 - 5 + 10x\\
\Leftrightarrow 8x = - 61\\
\Leftrightarrow x = - \dfrac{{61}}{8}\\
b)\dfrac{{3x - 2}}{6} - 5 = \dfrac{{3 - 2\left( {x + 7} \right)}}{4}\\
\Leftrightarrow 2\left( {3x - 2} \right) - 60 = 3\left( { - 11 - 2x} \right)\\
\Leftrightarrow 6x - 4 - 60 = - 33 - 6x\\
\Leftrightarrow 12x = 31\\
\Leftrightarrow x = \dfrac{{31}}{{12}}
\) \(a.\frac{x-3}{5}=6-\frac{1-2x}{2}\\\Leftrightarrow
\frac{2\left(x-3\right)}{10}=\frac{60}{10}-\frac{5\left(1-2x\right)}{10}\\
\Leftrightarrow2\left(x-3\right)=60-5\left(1-2x\right)\\\Leftrightarrow
2x-6=60-5+10x\\\Leftrightarrow
2x-10x=6+60-5\\\Leftrightarrow
-8x=61\\
\Leftrightarrow x=-\frac{61}{8}\) Vậy nghiệm của phương trình trên là \(-\frac{61}{8}\)
K
Khách
CM
Cao Minh Tâm
22 tháng 10 2018
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