\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)

\(=\dfrac{4.4^5.6.6^5}{3.3^5.2.2^5}\)

\(=\dfrac{4^6.6^6}{3^6.2^6}\)

\(=\dfrac{2^6.2^6.2^6.3^6}{3^6.2^6}\)

\(=2^{12}=2^{3^4}=8^4=8^x\)

Vậy x = 4

Vậy x = 4

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\) 

\(\Rightarrow\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) 

\(\Rightarrow\dfrac{4^5.4.6^5.6}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{\left(2.2\right)^5.2.2.\left(3.2\right)^5.3.2}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{2^5.2^5.2.2.3^5.2^5.3.2}{3^5.3.2^5.2}=2^n\) 

Rút gọn vế trái ta có :

\(2^5.2.2.^5=2^n\)

\(\Rightarrow2^{12}=2^n\) 

\(\Rightarrow n=12\) ( Thỏa mãn điều kiện \(n\in N\) ) 

Vậy n =12 

Lời giải:

\(\text{VT}=\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)

\(=\frac{4.4^5}{3.3^5}.\frac{6.6^5}{2.2^5}=\frac{4^6.6^6}{3^6.2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)

Do đó: \(8^{|2x+6|}=2^{12}\Leftrightarrow 2^{3|2x+6|}=2^{12}\)

\(\Leftrightarrow 3|2x+6|=12\Leftrightarrow |2x+6|=4\)

\(\Rightarrow\left[{}\begin{matrix}2x+6=4\\2x+6=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

\(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)

\(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{49}{30}\)

\(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}\)

\(\dfrac{8}{5}x\dfrac{5}{8}=\dfrac{1}{1}=1\)

\(\dfrac{6}{7}x\dfrac{4}{7}=\dfrac{24}{49}\)

\(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}x\dfrac{5}{4}=\dfrac{1}{1}=1\)

\(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}x\dfrac{5}{5}=\dfrac{1}{1}=1\)

1) \(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{1+2}{3}=\dfrac{3}{3}=1\)

2) \(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{24+25}{30}=\dfrac{49}{30}\)

3) \(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{4-3}{5}=\dfrac{1}{5}\)

4) \(\dfrac{9}{8}-\dfrac{4}{2}=\dfrac{9}{8}-2=\dfrac{9}{8}-\dfrac{16}{8}=-\dfrac{7}{8}\)

5) \(\dfrac{8}{5}\times\dfrac{5}{8}=\dfrac{8\times5}{5\times8}=\dfrac{40}{40}=1\)

6) \(\dfrac{6}{7}\times\dfrac{4}{7}=\dfrac{6\times4}{7}=\dfrac{24}{7}\)

7) \(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{5}{4}=\dfrac{4\times5}{5\times4}=\dfrac{20}{20}=1\)

8) \(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}\times\dfrac{5}{5}=\dfrac{5\times5}{5\times5}=\dfrac{25}{25}=1\)

giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều

a: =>1/2x-3/4x=-5/6+7/3

=>-1/4x=14/6-5/6=3/2

=>x=-3/2*4=-6

b: =>4/5x-3/2x=1/2+6/5

=>-7/10x=17/10

=>x=-17/7

c: =>6/5x+6/20=6/5-1/3x

=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10

=>x=27/46

d: =>6x+3/2+4/5=1/2-2x

=>8x=1/2-3/2-4/5=-1-4/5=-9/5

=>x=-9/40

a) \(2^{3x+2}=4^{x+5}\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\Leftrightarrow2^{3x+2}=2^{2x+10}\)

\(\Rightarrow3x+2=2x+10\Leftrightarrow3x+2-2x-10\)

\(\Leftrightarrow x-8=0\Leftrightarrow x=8\) vậy \(x=8\)

câu : b ; c mk cảm thấy đề sai

\(\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

\(\dfrac{5}{9}\times3=\dfrac{5}{9}\times\dfrac{3}{1}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

\(\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

\(\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

\(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

\(\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

\(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3\times1}{5\times2}=\dfrac{3}{10}\)

\(\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

a) \(\dfrac{3}{7}+4=\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

b) \(\dfrac{5}{9}\times3=\dfrac{5\times3}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

c) \(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

d) \(5-\dfrac{3}{4}=\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

e) \(\dfrac{5}{7}:6=\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

f) \(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

g) \(3+\dfrac{3}{4}=\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

h) \(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3\times4}{5\times8}=\dfrac{12}{40}=\dfrac{3}{10}\)

i) \(5:\dfrac{3}{8}=\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

=>\(\dfrac{4^5\left(1+1+1+1\right)}{3^5\left(1+1+1\right)}.\dfrac{6^5\left(1+1+1+1+1+1\right)}{2^5\left(1+1\right)}=2^n\)

=>\(\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) =>\(\dfrac{4^6}{3^6}.\dfrac{6^6}{2^6}=2^n\)

=>\(\left(\dfrac{4.6}{3.2}\right)^6=2^n\) =>\(4^6=2^n\) =>\(2^{12}=2^n\) =>n=12.

\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)

k/

\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)

\(\Leftrightarrow96-4x+8=3x\)

\(\Leftrightarrow96-4x+8-3x=0\)

\(\Leftrightarrow104-7x=0\)

\(\Leftrightarrow7x=104\)

\(\Leftrightarrow x=104:7\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)

m/ 

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

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1-12x-10=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)

k) Ta có: \(8-\dfrac{x-2}{2}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{32}{4}-\dfrac{2\left(x-2\right)}{4}=\dfrac{x}{4}\)

\(\Leftrightarrow32-2x+4-x=0\)

\(\Leftrightarrow28-x=0\)

hay x=28

Vậy: S={28}

m) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

\(\Leftrightarrow6x+5-12x-10=0\)

\(\Leftrightarrow-6x=5\)

hay \(x=-\dfrac{5}{6}\)

Vậy: \(S=\left\{-\dfrac{5}{6}\right\}\)

n) Ta có: \(\dfrac{x+1}{7}+\dfrac{x+2}{6}=\dfrac{x+3}{5}+\dfrac{x+4}{4}\)

\(\Leftrightarrow\dfrac{x+1}{7}+1+\dfrac{x+2}{6}+1=\dfrac{x+3}{5}+1+\dfrac{x+4}{4}+1\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}=\dfrac{x+8}{5}+\dfrac{x+8}{4}\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}-\dfrac{x+8}{4}=0\)

\(\Leftrightarrow\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

mà \(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)

nên x+8=0

hay x=-8

Vậy: S={-8}