a) \(\left(x-1\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)

b) \(\left(x^2+5\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)

mà \(x\in Z\Rightarrow x=5\)

c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)

mà \(x\in Z\Rightarrow x\in\varnothing\)

\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

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

1.

a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)

b) x=0

d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)

e) \(x=\frac{2}{3}\)

a)\(\left(\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^4.2^3=3-1+\frac{1}{16}.8=3-1+\frac{1}{2}=\frac{5}{2}\\ \)

b)\(2^2.2^3.\left(\frac{2}{3}\right)^{-2}=2^5.\frac{9}{4}=72\)

c)\(\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^5:\left(\frac{3}{2}\right)^3=\frac{9}{128}\)

2)

\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)

\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)

\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)

a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)

d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)

e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)

g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)

h)Tương tự các câu trên

i) x = 0

k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)

l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)

=> x + 1 = 3 => x = 2

x.(x+1)=0

suy ra x=0 hoac x+1=0

                               x=0-1

                              x=-1

vay x=0 hoac  x=-1

mấy câu sau cũng làm tương tự

\(a,\left(x+1\right)^2=81\) 

    \(\left(x+1\right)^2=9^2\)  Hoặc \(\left(x+1\right)^2=\left(-9\right)^2\)

      \(\left(x+1\right)=9\)                     \(x+1=-9\)

                     \(x=8\)                               \(x=-10\)

b,\(\left(x+5\right)^{^{ }3}=-64\)

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

          \(x+5=-4\)

=>               \(x=-9\)

c,\(\left(2x-3\right)^2=9\)

=>\(\left(2x-3\right)^2=3^2\)Hoặc  \(\left(2x-3\right)^2=\left(-3\right)^2\)

            \(2x-3=3\)                    \(2x-3=-3\)

                     \(2x=6\)                             \(2x=0\)       

=> \(\hept{\begin{cases}x=3\\x=0\end{cases}}\)

d, \(\left(4x+1\right)^3=27\)

   \(\left(4x+1\right)^{^{ }3}=3^3\)

            \(4x+1=3\)

                     \(4x=2\)

                       \(x=\frac{1}{2}\)

\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{8^6}{4}=\frac{\left(2^3\right)^6}{2^2}=\frac{2^{18}}{2^2}=2^{16}\)

\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{15}+4^{10}}{4^6+4^{11}}=\frac{4^{10}.4^5+4^{10}}{4^6+4^6.4^5}=\frac{4^{10}.\left(4^5+1\right)}{4^6.\left(4^5+1\right)}=\frac{4^{10}}{4^6}=4^4=256\)

phần D trên mk làm sai xin lỗi nha