(1/16)^x+2=(1/32)^6
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\dfrac{1}{16}\right)^{x+2}=\left(\dfrac{1}{32}\right)^6\)
\(\Leftrightarrow\dfrac{1}{16^{x+2}}=\dfrac{1}{32^6}\)
\(\Leftrightarrow\dfrac{1}{2^{4x+8}}=\dfrac{1}{2^{30}}\)
\(\Leftrightarrow4x+8=30\Leftrightarrow4x=22\Leftrightarrow x=\dfrac{11}{2}\)
3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
6:=(3/2)*(3/2)^2*(3/2)^4=(3/2)^7
7: =(1/2)^7*2^3*2^5*2^8=2^9
8: =(-1/7)^4*5^4=(-5/7)^4
9: =2^2*2^5:(2^3/2^4)
=2^7/2=2^6
10: =(1/7)^3*7^2=1/7
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
\(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)
\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8=\left(\dfrac{1}{7}\right)^7\times2^{16}\)
\(\left(-\dfrac{1}{7}\right)^4\times125\times5=\left(-\dfrac{1}{7}\right)^4\times5^3\times5=\left(-\dfrac{1}{7}\right)^4\times5^4\)
\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:2^3:2^{-4}=2^0\)
\(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^3\times7^3=1^3\)
6, \(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4\)
7,\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8\)
8,\(\left(-\dfrac{1}{7}\right) ^4\times125\times5=\left(\dfrac{1}{7}\right)^4\times5^3\times5\)
9,\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:\left[2^3\times\left(\dfrac{1}{2}\right)^4\right]\)
10, \(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times7^2\)
Bài làm :
\(\left(\frac{1}{16}\right)^{x+2}=\left(\frac{1}{32}\right)^6\)
\(\Rightarrow\left(\frac{1}{2^4}\right)^{x+2}=\left(\frac{1}{2^5}\right)^6\)
\(\Rightarrow\left(\frac{1}{2}\right)^{4x+8}=\left(\frac{1}{2}\right)^{30}\)
\(\Rightarrow4x+8=30\)
\(\Rightarrow4x=22\)
\(\Rightarrow x=\frac{11}{2}\)
Học tốt nhé
\(\left(\frac{1}{16}\right)^{x+2}=\left(\frac{1}{32}\right)^6=>\left[\left(\frac{1}{2}\right)^4\right]^{x+2}=\left[\left(\frac{1}{2}\right)^4\right]^6\)
\(\left(\frac{1}{2}\right)^{4.\left(x+2\right)}=\left(\frac{1}{2}\right)^{30}\)
4.(x+2)=30
(x+2)=30:4
(x+2)=7,5
x=7,5+2
x=9,5
vậy x=9,5