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\(\frac{1.2.3.4....30.31}{2.2.2.3.2.3.....2.32}=\frac{2.3.4....30.31}{2^{31}\left(2.3...31\right).32}=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^{-36}\)
Vậy x=-36
ta có \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}.....\frac{30}{62}\cdot\frac{31}{64}=2^x\)
=>\(\frac{1.2.3.4....31}{2\cdot2\cdot2\cdot3\cdot2\cdot3.....\cdot2\cdot3\cdot2}=\frac{2\cdot3\cdot4...30.31}{2^{31}\left(2\cdot3\cdot4...31\right)32}=\frac{1}{2^{31}\cdot2^5}=\frac{1}{2^{36}}=2^{-36}\)
\(=>x=-36\)
\(=\frac{1x2x3x...x30x31}{2^{31}x\left(2x3x4x...x31x32\right)}=\frac{1}{2^{31}x32}=\frac{1}{2^{36}}\)
1/4 . 2/6 . 3/8 . ... .30/62 .31/64 = 2^x
(1/2 . 1/2).(2/3 . 1/2).(3/4 . 1/2). ... .(30/31 . 1/2).(31/32 . 1/2) = 2^x
(1/2.1/2. ... .1/2).(1/2 . 2/3 . 3/4. ... .30/31 . 31/32) = 2^x
(31 số 1/2)
(1/2)^31. = 2^x
=> 0=x+36
x=0-36
x=-36
Vậy x=-36
Theo mk nghĩ,mk làm đúng nha .Tk cho mk
Để mk sửa phần này một chút
\((\frac{1}{2})^{31}\cdot\frac{1\cdot2\cdot3.....30\cdot31}{2\cdot3\cdot4.....31\cdot32}=2^x\)
\(\frac{1^{31}}{2^{31}}\cdot\frac{1}{32}=2^x\)
\(\frac{1}{2^{31}}\cdot\frac{1}{2^5}=2^x\)
\(\frac{1}{2^{36}}=2^x\)
\(1=2^x\cdot2^{36}\)
\(2^0=2^x+36\)
Rồi bn tự suy luận nha
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)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
Có thể đề là: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.....\frac{31}{64}=2^n\)
=> \(\frac{1.2.3.4....31}{\left(2.2\right)\left(2.3\right).\left(2.3\right)\left(2.4\right)\left(2.5\right)...\left(2.31\right).\left(2.32\right)}=2^n\)
=> \(\frac{1.2.3.4...31}{2^{16}.\left(2.3.4.5..31.32\right)}=2^n\) => \(\frac{1}{2^{16}.32}=2^n\) => 216.25.2n = 1
=> 231+n = 1 = 20 => 31 + n = 0 => n = -31
\(=\frac{1.2.3.....30}{4.6.8....64}=\frac{1}{2.2...2.64}=\frac{1}{2^{30}.2^6}=\frac{1}{2^{36}}\) ( 30 số 2)
=> 2^n = 1/2^36
=> n = -36
\(P=...\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)
\(=\frac{1}{99}-1=\frac{-98}{99}\)
\(M=...\)
\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)
\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)
\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)
\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)
\(\frac{4}{1}.\frac{6}{2}.\frac{8}{3}.\frac{10}{4}...\frac{64}{31}=\frac{2^{31}.\left(2.3.4.5...32\right)}{1.2.3.4...31}=2^{31}.32\)
Mà \(2^x=2^{31}.32=2^{31}.2^5=2^{36}\)
\(\Rightarrow x=36\)
\(E=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=\frac{1\cdot1\cdot1\cdot1\cdot.....\cdot1\cdot1}{2\cdot2\cdot2\cdot....\cdot2\cdot64}=\frac{1}{2\cdot30\cdot64}=\frac{1}{3840}\)