b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)

                                                                                                       \(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)

   Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)                                                    

Vậy x= 4

a.4^7

b.8^5 

 c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu

\(1,\)\(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)

        \(x-\frac{3}{5}=\frac{3}{35}+\frac{7}{6}\)

        \(x-\frac{3}{5}=\frac{263}{210}\)

                    \(x=\frac{263}{210}+\frac{3}{5}\)

                    \(x=\frac{389}{210}\)

        VẬY: \(x=\frac{389}{210}\)

1. \(x=\frac{61}{42}\)

2. \(x=\frac{-36}{5}\) 

3. \(x=\frac{13}{11}\)

4. \(x=\frac{1}{12}\)

5.\(x=\frac{-5}{2}\)

a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)

Do đó \(x\in\left\{0;1;2\right\}\)

b)

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)

\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)

 o0oNguyễno0o bn giúp mk nha bài này khó wa

 Harry Potter05 giúp mk đc ko vậy bn!!!

 o0oNguyễno0o giúp mk nhanh ik nha

Bài 1 : Thực hiện phép tính :

a, \(\frac{4}{5}+1\frac{1}{6}\cdot\frac{3}{4}\)

= \(\frac{4}{5}+\frac{7}{6}\cdot\frac{3}{4}\)

= \(\frac{4}{5}+\frac{7}{8}\)

= \(\frac{32+35}{40}=\frac{67}{40}\)

b, \(\frac{2}{3}:\left(\frac{3}{4}\cdot\frac{4}{3}\right)+2\)

\(=\frac{2}{3}:1+2\)

\(=\frac{2}{3}+2=\frac{2+6}{3}=\frac{8}{3}\)

c, \(\frac{1}{2}\times\left(\frac{2}{3}+\frac{3}{5}\cdot\frac{5}{7}\right)+1\frac{1}{3}\)

\(=\frac{1}{2}\cdot\left(\frac{2}{3}+\frac{9}{35}\right)+\frac{4}{3}\)

\(=\frac{1}{2}\cdot\frac{97}{105}+\frac{4}{3}\)

\(=\frac{97}{210}+\frac{4}{3}=\frac{377}{210}\)

Bài 2 : Tìm \(x\inℤ\), biết :

a, \(\frac{2}{3}< \frac{x}{6}\le\frac{10}{3}\)

\(\Leftrightarrow\frac{4}{6}< \frac{x}{6}\le\frac{20}{6}\)

mà \(x\inℤ\Rightarrow\text{x}\in\) {\(5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20\)}

b, \(\frac{1}{3}+x=1\frac{1}{2}\)

\(\frac{1}{3}+x=\frac{3}{2}\)

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

\(x=\frac{7}{6}\) (loại vì \(x\notinℤ\))

\(\Rightarrow x\in\varnothing\)

c, \(\frac{1}{7}+x=\frac{25}{14}+\frac{5}{14}\)

\(\frac{1}{7}+x=\frac{15}{7}\)

\(x=\frac{15}{7}+\frac{(-1)}{7}\)

\(x=\frac{14}{7}=2\).