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24 tháng 7 2019

a) \(3x-\frac{1}{5}=\frac{4+x}{2}\)

=> \(\frac{15x-1}{5}=\frac{4+x}{2}\)

=> \(\left(15x-1\right).2=\left(4+x\right).5\)

=> \(30x-2=20+5x\)

=> \(30x-5x=20+2\)

=> \(25x=22\)

=>  \(x=\frac{22}{25}\)

b) \(\frac{4}{3}x-1=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)

=> \(\frac{1}{3}x=\frac{4x+4-1}{3}\)

=> \(\frac{1}{3}x=\frac{4x+3}{3}\)

=> \(3x=3\left(4x+3\right)\)

=> \(3x=12x+9\)

=> \(3x-12x=9\)

=> \(-9x=9\)

=> \(x=9:\left(-9\right)=-1\)

24 tháng 7 2019

c đâu bạn

16 tháng 8 2019

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

16 tháng 8 2019

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

5 tháng 9 2016

a)|x+0,573|=2

=>x+0,573=2 hoặc -2

Xét x+0,573=2

=>x=1,427

Xét x+0,573=-2

=>x=-2,573

 

5 tháng 9 2016

a) | x + 0,573 | = 2

\(\Rightarrow\)x + 0,573 = 2 hoặc x + 0,573 = -2

+) x + 0,573 = 2\(\Rightarrow\)x = 1,427

+) x + 0,573 = -2\(\Rightarrow\)x = -2,573

Vậy x = 1,427 hoặc -2,573

b) \(\left|x+\frac{1}{3}\right|-4=-1\)

\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)

\(\Rightarrow x+\frac{1}{3}=3\) hoặc \(x+\frac{1}{3}=-3\)

+) \(x+\frac{1}{3}=3\Rightarrow x=\frac{8}{3}\)

+) \(x+\frac{1}{3}=-3\Rightarrow x=\frac{-10}{3}\)

Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{-10}{3}\)

Các phần khác làm tương tự nhé bạn

3 tháng 7 2019

a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)

=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)

=> \(\left|2-\frac{3}{2}x\right|=x+6\)

ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)

Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)

=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)

=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)

b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)

=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)

=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)

=> x = 1/4

hoặc x = 0 hoặc x = 1/2

26 tháng 10 2016

a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)

        \(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)

        \(\frac{2}{5}-x=-3\)

                   \(x=\frac{2}{5}-\left(-3\right)\)

                   \(x=\frac{2}{5}+3\)

                   \(x=\frac{3}{5}-\frac{15}{5}\)

                   \(x=-\frac{12}{5}\)

Vay \(x=-\frac{12}{5}\) 

    

  

26 tháng 10 2016

b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)

        \(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)

        \(-3+\frac{3}{x}=\frac{-25}{12}\)

                     \(\frac{3}{x}=\frac{-25}{12}+3\)

                      \(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)

                      \(\frac{3}{x}=\frac{5}{6}\)

                      \(\frac{18}{6x}=\frac{5x}{6x}\)

Đèn dây , bạn tự làm tiếp nhé , de rồi chứ

19 tháng 4 2020
https://i.imgur.com/wgXaoMx.jpg

b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)

\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)

\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)

\(9x-10=0\)

hay 9x=10

\(x=\frac{10}{9}\)

Vậy: \(x=\frac{10}{9}\)

c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)

\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)

\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)

\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)

\(6x-3-5x+10-3x-21=0\)

\(-2x-14=0\)

\(-2x=14\)

hay x=-7

Vậy: x=-7

d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)

\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

\(6x-18+7x-35-13x-4=0\)

\(-21\ne0\)

Vậy: x∈∅

e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)

\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)

\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)

\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)

\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)

\(-12x+96=0\)

\(-12x=-96\)

hay x=8

Vậy: x=8

19 tháng 2 2018

      \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow\)\(x+329=0\)   (vì  1/327 + 1/326 + 1/325 + 1/324 + 1/5  khác  0  )

\(\Leftrightarrow\)\(x=-329\)

19 tháng 2 2018

Bài 1 : 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

\(\Rightarrow\)\(x+329=0\)

\(\Rightarrow\)\(x=-329\)

Vậy \(x=-329\)

18 tháng 10 2018

\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)

\(=\frac{7}{2}-2\)

\(=\frac{7}{2}-\frac{4}{2}\)

\(=\frac{3}{2}\)

\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)

\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)

\(=\frac{3}{7}.\left(2-9\right)\)

\(=\frac{3}{7}.\left(-7\right)\)

\(=-3\)

\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )

18 tháng 10 2018

a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)

\(3\cdot25:\frac{5}{4}\)

\(3\cdot\left(25:\frac{5}{4}\right)\)

=\(3\cdot20\)

=60

b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)

=\(\frac{3}{7}\cdot\left(-7\right)\)

=\(-3\)

c) =