a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)

\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)

=>x=0

c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)

d: \(\Leftrightarrow x^8=x^7\)

=>x(x-1)=0

=>x=0(loại) hoặc x=1(nhận)

e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)

hay x=1

f: =>x-1=20

hay x=21

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

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

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

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

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

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

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

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

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

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

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

\(=-3\)

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

1 

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) = 

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

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

cho mình

Giải phương trình

a) \(\frac{2x}{x-1}-\frac{x}{x-2}=\frac{x^2}{\left(x-1\right)\left(x-2\right)}\left(x\ne1,x\ne2\right)\)

\(\Leftrightarrow\frac{2x\left(x-2\right)-x\left(x-1\right)-x^2}{\left(x-1\right)\left(x-2\right)}=0\)

\(\Rightarrow2x^2-x^2-x^2-4x+x=0\)

\(\Leftrightarrow-3x=0\Leftrightarrow x=0\left(tm\right)\)

KL: Vậy...

b)\(\frac{1}{x+2}-\frac{6}{x-1}+\frac{8}{\left(x+2\right)\left(x-1\right)}=0\left(x\ne-2,x\ne1\right)\)

\(\Leftrightarrow\frac{\left(x-1\right)-6\left(x+2\right)+8}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Rightarrow x-1-6x-12+8=0\)

\(\Leftrightarrow-5x=-7\Leftrightarrow x=\frac{7}{5}\left(tm\right)\)

c) \(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x+3\right)\left(x-1\right)}\left(x\ne-3,x\ne1\right)\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)-\left(x+1\right)\left(x+3\right)-4}{\left(x+3\right)\left(x-1\right)}=0\)

\(\Rightarrow x^2+x-2-x^2-4x-3-4=0\)

\(\Leftrightarrow-3x=9\Leftrightarrow x=-3\left(ktm\right)\)

a)\(\frac{5}{6}-x=-\frac{7}{12}+\frac{2}{3}\)

     \(\frac{5}{6}-x=\frac{1}{12}\)

      \(x=\frac{5}{6}-\frac{1}{12}\)

      \(\Rightarrow x=\frac{3}{4}\)

b)\(\left(2,4x-36\right):1\frac{5}{7}=-14\)

    \(\left(2,4x-36\right)=-24\)

      \(2,4x=12\)

       \(\Rightarrow x=5\)

c)\(\left(3\frac{1}{2}+2x\right).3\frac{2}{3}=5\frac{1}{3}\)

    \(3\frac{1}{2}+2x=\frac{16}{11}\)

     \(2x=-\frac{45}{22}\)

      \(x=-\frac{45}{44}\)

d)\(\frac{5}{6}-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{3}{8}\)

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{11}{24}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{11}{24}\\\frac{1}{2}x-\frac{1}{3}=-\frac{11}{24}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{19}{12}\\x=-\frac{1}{4}\end{cases}}\)

e)\(\left|\frac{1}{4}-2x\right|-\frac{3}{4}=0\)

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

    \(\Rightarrow\hept{\begin{cases}\frac{1}{4}-2x=\frac{3}{4}\\\frac{1}{4}-2x=-\frac{3}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{2}\end{cases}}\)

f)\(\left|x-\frac{1}{3}\right|+\frac{2}{3}=0\)

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

           Vì trị tuyệt đối  ko âm nên PT vô nghiệm

g)\(\left(\frac{3}{15}-x\right).\frac{1}{3}=\frac{2}{5}\)

      \(\frac{3}{15}-x=\frac{6}{5}\)

       \(x=\frac{3}{25}-\frac{6}{5}\)

      \(x=-\frac{27}{25}\)

a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)

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

\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)

\(\Rightarrow x+3=-3\)

\(\Rightarrow x=-6\)

b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)

\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)

\(\Rightarrow2x+2=-2\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)

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

\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=-9\)

d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)

\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)

\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

f)\(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)

\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)

\(5^{x+3}\left(5-3\right)=2\cdot5^{11}\)

\(5^{x+3}\cdot2=2\cdot5^{11}\)

\(\Rightarrow5^{x+3}=5^{11}\)

\(\Rightarrow x+3=11\)

\(\Rightarrow x=8\)

r)\(4\cdot3^{x-1}+2\cdot3^{x+2}=4\cdot3^6+2\cdot3^9\)

\(4\cdot3^x:3+2\cdot3^x\cdot9=4.3^7:3+2\cdot3^7\cdot9\)

\(3^x\left(4:3+2\cdot9\right)=3^7\left(4:3+2\cdot9\right)\)

\(\Rightarrow3^x=3^7\)

\(\Rightarrow x=7\)