Ta có : \(\frac{3}{2}.4^x+\frac{5}{4}.4^{x+2}=\frac{3}{2}.4^8+\frac{5}{3}.4^{10}\)

\(\Rightarrow4^x\left(\frac{3}{2}+\frac{5}{3}.4^2\right)\)=\(4^8\left(\frac{3}{2}+\frac{5}{3}.4^2\right)\)

\(\Rightarrow4^x=4^8\)

\(\Rightarrow x=8\)

\(Can\) \(you\) \(k\)\(for\)\(me?\)

Dễ lắm

trả lời đi

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)

hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)

Vậy: \(x=\frac{9}{10}\)

b) Ta có: \(5\frac{4}{7}:x=13\)

\(\Leftrightarrow\frac{39}{7}:x=13\)

\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)

Vậy: \(x=\frac{3}{7}\)

c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=84\)

\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)

Vậy: x=30

d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)

hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)

Vậy: x=-5

e) Ta có: \(8\frac{2}{3}:x-10=-8\)

\(\Leftrightarrow\frac{26}{3}:x=2\)

hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)

Vậy: \(x=\frac{13}{3}\)

g) Ta có: \(x+30\%=-1.3\)

\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)

hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)

Vậy: \(x=\frac{-8}{5}\)

i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)

\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)

\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)

\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)

Vậy: x=-9

k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)

hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)

Vậy: x=30

m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)

\(\Leftrightarrow\left|2x-1\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)

thank you nha!

\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)

\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)

\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)