Tìm x
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{9}{10}.....\frac{36}{62}.\frac{31}{64}=\frac{1}{2^x}\)
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\(\frac{1.2.3....31}{2^{30}.\left(2.3....31\right).32}=\frac{1}{2^{31}.32}=\frac{1}{2^{36}}=2^{-36}=2^x\)
Vậy x=-36
Hok tốt
\(\frac{1.2.3.4....30.31}{2.2.2.3.2.3.....2.32}=\frac{2.3.4....30.31}{2^{31}\left(2.3...31\right).32}=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^{-36}\)
Vậy x=-36
ta có \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}.....\frac{30}{62}\cdot\frac{31}{64}=2^x\)
=>\(\frac{1.2.3.4....31}{2\cdot2\cdot2\cdot3\cdot2\cdot3.....\cdot2\cdot3\cdot2}=\frac{2\cdot3\cdot4...30.31}{2^{31}\left(2\cdot3\cdot4...31\right)32}=\frac{1}{2^{31}\cdot2^5}=\frac{1}{2^{36}}=2^{-36}\)
\(=>x=-36\)
\(\Rightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}...\frac{30}{2.31}.\frac{31}{2.32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}}.\frac{1.2.3.4....31}{2.3.4...32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}}.\frac{1}{32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Rightarrow\frac{1}{2^{36}}=2^x\Rightarrow x=-36\)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot......\cdot\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot31}{4\cdot6\cdot8\cdot....\cdot64}=2^x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot31}{\left(2\cdot2\right)\cdot\left(3\cdot2\right)\cdot\left(4\cdot2\right)\cdot.....\cdot\left(2\cdot32\right)}=2^x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{\left(2\cdot2\cdot2\cdot....\cdot2\right)\left(1\cdot2\cdot3\cdot.....\cdot31\right)\cdot32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=2^x\)
=>\(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}....\frac{30}{2.31}.\frac{31}{2.32}=2^x\)
=>\(\frac{1.2.3.4.5....30.31}{2.2.2.3.2.4.2.5.2.6...2.31.2.32}=2^x\)
=>\(\frac{2.3.4.5...30.31}{2^{31}.32.\left(2.3.4.5...31\right)}=2^x\)
=>\(\frac{1}{2^{31}.2^5}=2^x\)
=>\(\frac{1}{2^{36}}=2^x\)
=> x=36
Vậy x=36
Chúc bn học tốt nhé!
Ta có: \(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}.......\frac{30}{2.31}.\frac{31}{64}=4^x\)
\(\frac{1}{2^{30}.64}=4^x\Leftrightarrow4^x.2^{30}.2^6=1\)
\(\Leftrightarrow2^{2x+36}2^0\)
\(\Leftrightarrow2x+36=0\)
\(\Leftrightarrow2x=-36\)
\(\Leftrightarrow x=-18\)
Vậy ........
$4^x.64=1$\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=4^x\)
\(\Leftrightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}.....\frac{30}{2.31}.\frac{31}{2.32}=4^x\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.....\frac{30}{31}.\frac{31}{32}\right)=4^x\)
\(\Leftrightarrow\frac{1}{2}.\frac{1.2.3.4.5.....30.31}{2.3.4.5.6.....31.32}=4^x\)
\(\Leftrightarrow\frac{1}{2}.\frac{1}{32}=4^x\)
\(\Leftrightarrow4^x=\frac{1}{64}\)
\(\Leftrightarrow4^x.64=1\)
\(\Leftrightarrow4^x.4^3=1\Leftrightarrow4^{x+3}=4^0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy x = -3
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