\(\dfrac{2^{2x-3}}{4^{10}}=8^3.16^5\)

\(\Rightarrow2^{2x-3}=8^3.16^5.4^{10}\)

\(\Rightarrow2^{2x-3}=2^9.2^{20}.2^{20}\)

\(\Rightarrow2^{2x-3}=2^{49}\)

Vì \(2\ne\pm1;2\ne0\) nên \(2x-3=49\)

\(\Rightarrow2x=52\Rightarrow x=26\)

Vậy..........

Chúc bạn học tốt!!!

a: \(\dfrac{5^5}{5^x}=5^{18}\)

=>5-x=18

hay x=-13

b: \(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Leftrightarrow2^{4-x}=\left(2^5\right)^6\cdot\left(2^4\right)^5=2^{30+20}=2^{50}\)

=>4-x=50

hay x=-46

c: \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)

\(\Leftrightarrow2^{2x-3}=2^9\cdot2^{20}\cdot2^{20}=2^{49}\)

=>2x-3=49

=>2x=52

hay x=26

d: \(\dfrac{2^3}{2^x}=4^5\)

\(\Leftrightarrow2^{3-x}=2^{10}\)

=>3-x=10

hay x=-7

e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)

\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)

\(\Leftrightarrow5^x=5^6\)

hay x=6

f: \(7\cdot2^x=2^9+5\cdot2^8\)

\(\Leftrightarrow2^x\cdot7=2^8\cdot7\)

\(\Leftrightarrow2^x=2^8\)

hay x=8

a, Tự chép đề bài :v

=> 2^2x-3 = ( 8³. 16⁵ ) : 4¹⁰

2^2x-3 = ( 2⁹. 2²⁰ ) : 2²⁰

2^2x-3 = 2²⁹ : 2²⁰

2^2x-3 = 2⁹

=> 2x - 3 = 9

2x = 9 + 3

2x = 12

x = 6 

Vậy....

b, 7. 2^x = 2⁹ + 5. 2⁸

7. 2^x = 1792

2^x = 1792 : 7

2^x = 256

2^x = 2⁸

=> x = 8 

Vậy ....

a  \(\frac{2^{2x-3}}{4^{10}}=8^3.16^5\)

\(\Leftrightarrow\frac{2^{2x-3}}{4^{10}}=2^{29}\)

\(\Leftrightarrow2^{2x-3}=2^{29}.4^{10}\)

\(\Leftrightarrow2^{2x-3}=2^{49}\)

\(\Leftrightarrow2x-3=49\)

\(\Leftrightarrow x=26\)

b  \(7.2^x=2^9+5.2^8\)

\(\Leftrightarrow7.2^x=2^8.(2+5)\)

\(\Leftrightarrow7.2^x=7.2^8\)

\(\Leftrightarrow x=7\)

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

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

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

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

2:

a: x=2,4-0,4=2

b: =>2x=-1,5+0,8=-0,7

=>x=-0,35

c: =>x-16=-15

=>x=1

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

bn ơi ktra lại câu 2 giúp mk đc ko 

a) \(\dfrac{-2}{3}+2x=\dfrac{4}{3}\)

\(\Rightarrow2x=\dfrac{4}{3}-\dfrac{-2}{3}\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

b) \(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)

\(\Rightarrow-5:x=\dfrac{5}{8}-\dfrac{-3}{8}\)

\(\Rightarrow-5:x=1\)

\(\Rightarrow x=-5\)

a) \(\dfrac{-2}{3}+2x=\dfrac{4}{3}\)

\(\Rightarrow2x=\dfrac{4}{3}-\dfrac{-2}{3}\)

\(\Rightarrow2x=\dfrac{6}{3}\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=\dfrac{2}{2}\)

\(\Rightarrow x=1\)

b) \(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)

\(\Rightarrow-5:x=\dfrac{-3}{8}-\dfrac{5}{8}\)

\(\Rightarrow-5:x=-\dfrac{8}{8}\)

\(\Rightarrow-5:x=-1\)

\(\Rightarrow x=-5:-1\)

\(\Rightarrow x=5\)