1/4.2/6.3/8.4/10.........30/62.31/64=4^x

^{=1/2.1/2.1/2.1/2.............1/2.1/64=4^x}

^{=1/2^30.1/2^6=4^x}

^{=1/2^36=4^x}

^{=1/4^18=4^x}

^=>x=-18

\(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}.....\dfrac{30}{62}.\dfrac{31}{64}=2^x\)

\(\Leftrightarrow\dfrac{1}{2.2}.\dfrac{2}{2.3}.\dfrac{3}{2.4}.\dfrac{4}{2.5}.\dfrac{5}{2.6}.....\dfrac{30}{2.31}.\dfrac{31}{2.32}=2^x\)

\(\Leftrightarrow\dfrac{1.2.3.4.5.....30.31}{2.2.2.3.2.4.2.5.2.6.....2.31.2.32}=2^x\)

\(\Leftrightarrow\dfrac{2.3.4.5.....30.31}{2^{31}.32.\left(2.3.4.5.....31\right)}=2^x\)

\(\Leftrightarrow\dfrac{1}{2^{31}.2^5}=2^x\)

\(\Leftrightarrow\dfrac{1}{2^{36}}=2^x\)

\(\Leftrightarrow2^{-36}=2^x\)

\(\Leftrightarrow x=-36\)

https://hoc24.vn/hoi-dap/question/279983.html

vào link này nha

Quang Anh:" Mạnh giòi thì về search mạng chí mồ"

Mạnh :" cũng khó không biết chi im mồm"

Và kết quả là đây

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)

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

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

Bài 1:

a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)

\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)

\(=\frac{2+3}{12}=\frac{5}{12}\)

b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)

\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)

\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)

\(=\frac{23}{12}-\frac{-7}{10}\)

\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)

Bài 2:

a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)

\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)

\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)

Vậy: \(x=\frac{6}{5}\)

b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)

\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)

\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)

\(\Leftrightarrow-2x=-6-1=-7\)

hay \(x=\frac{7}{2}\)

Vậy: \(x=\frac{7}{2}\)

Bài 1:

a) Ta có: 

b) Ta có: 

Bài 2:

a) Ta có: 

Vậy: 

b) Ta có: 

hay 

Vậy: 

a.4^7

b.8^5 

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

>>>>>>>>>>>>>>>>>>/??????????????????????????

2^x nha

mình viết nhầm

b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

A = ( 1 - 1/2) . ( 1 - 1/3 ) . (1-1/4) ....(1-1/2015) . (1-1/2016)

A= 1/2 . 2/3 . 3/4...2014/2015 . 2015/2016

A = 1 . 2 . 3 . 4 ... 2014 . 2015/ 2 . 3 . 4 ... 2015 . 2016

A = 1/ 2016