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PT
1
2 tháng 9 2016
Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)
12 tháng 12 2019
\(2^x\times8^{x-1}=32\Leftrightarrow2^x\times2^{3x-3}=32\)
\(\Leftrightarrow2^{4x-3}=32\Leftrightarrow2^{4x-3}=2^5\)
\(\Rightarrow4x-3=5\Leftrightarrow4x=8\Rightarrow x=2\)
vậy \(x=2\)
12 tháng 12 2019
\(Ta có : 2 ^x . 8\)\(x - 1\) \(= 32\)
\(\Rightarrow\)\(2 ^ x . 2\)\(3. ( x - 1 )\) \(=\) \(32\)
\(\Rightarrow\)\(2 \)\(x + 3. ( x - 1 ) \)\(= 32\)
\(\Rightarrow\)\(2\)\(x + 3x - 3 \) \(= 32\)
\(\Rightarrow\)\(2\)\(4x - 3\) \(= 32\)
\(\Rightarrow\)\(2\)\(4x - 3\)\(= 2^5\)
\(\Rightarrow\)\(4x - 3 = 5\)
\(\Rightarrow\)\(4x = 8\)
\(\Rightarrow\)\(x = 2\)
\(Vậy : x = 2\)
T
16 tháng 9 2018
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}....\frac{15}{32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{2}{3}.\frac{1}{2}.\frac{3}{4}....\frac{1}{2}.\frac{15}{16}\)
\(=\left(\frac{1}{2}.\frac{1}{2}.....\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{15}{16}\right)\)
Số số \(\frac{1}{2}\)là (15-1):1+1=15 số
Do đó \(\left(\frac{1}{2}.\frac{1}{2}.....\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{15}{16}\right)=\left(\frac{1}{2}\right)^{15}.\frac{1.2.3....15}{2.3.4....15.16}\)
\(=\frac{1}{2^{15}}.\frac{1}{16}=\frac{1}{2^{15}}.\frac{1}{2^4}=\frac{1}{2^{15+4}}=\frac{1}{2^{19}}\)
\(\Rightarrow2^x=\frac{1}{2^{19}}\)
\(\Rightarrow x=-19\)
12 tháng 7 2019
a) \(\frac{x}{x+1}=\frac{1}{2}\)
=> 2x = x + 1
=> 2x - x = 1
=> x = 1
b) \(\frac{x}{2}=\frac{x}{3}\)
=> 3x = 2x
=> 3x - 2x = 0
=> x = 0
c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)
=> \(2017\left(x+1\right)=2\left(x+1\right)\)
=> 2017x + 2017 = 2x + 2
=> 2017x - 2x = 2 - 2017
=> 2015x = -2015
=> x = -2015 : 2015
=> x = -1
i) \(\frac{3}{x}=\frac{x}{2017}\)
=> x2 = 2017.3
=> x2 = 6051
=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)
còn lại tự lm
12 tháng 7 2019
\(a,\frac{x}{x+1}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow x=1\)
\(b,\frac{x}{2}=\frac{x}{3}\)
\(\Rightarrow x=\frac{x}{3}.2\)
\(\Rightarrow x=\frac{2x}{3}\)
\(\Rightarrow3x=2x\)
\(\Rightarrow x=0\)
\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)
\(\Rightarrow x+1=\frac{x+1}{2017}.2\)
\(\Rightarrow x+1=\frac{2x+2}{2017}\)
\(\Rightarrow2017x+2017=2x+2\)
\(\Rightarrow2017x-2x=2-2017\)
\(\Rightarrow2015x=-2015\)
\(\Rightarrow x=-1\)
\(i,\frac{3}{x}=\frac{x}{2017}\)
\(\Rightarrow x=3:\frac{x}{2017}\)
\(\Rightarrow x=\frac{6051}{x}\)
\(\Rightarrow x^2=6051\)
\(\Rightarrow x=\sqrt{6051}\)
\(o,\frac{x}{3}=\frac{x+1}{2}\)
\(\Rightarrow x=\frac{x+1}{2}.3\)
\(\Rightarrow x=\frac{3x+3}{2}\)
\(\Rightarrow2x=3x+3\)
\(\Rightarrow-x=3\)
\(\Rightarrow x=-3\)
\(m,\frac{x+1}{2}=\frac{x+2}{3}\)
\(\Rightarrow x+1=\frac{x+2}{3}.2\)
\(\Rightarrow x+1=\frac{2x+4}{3}\)
\(\Rightarrow3x+3=2x+4\)
\(\Rightarrow x=1\)
\(p,\frac{x+1}{2}=x\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow x=1\)
\(m,\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x=2:\frac{x}{8}\)
\(\Rightarrow x=\frac{16}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=4\)
\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)
\(\Rightarrow x^2=\frac{8}{x^2}.2\)
\(\Rightarrow x^2=\frac{16}{x^2}\)
\(\Rightarrow x^4=16\)
\(\Rightarrow x=2\)
\(r,\frac{x^3}{2}=\frac{32}{x}\)
\(\Rightarrow x^3=\frac{32}{x}.2\)
\(\Rightarrow x^3=\frac{64}{x}\)
\(\Rightarrow x^4=64\)
\(\Rightarrow x=\sqrt[4]{64}\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
13 tháng 12 2019
\(2^x.8^{x-1}=32\Leftrightarrow2^x=2^x.2^{3\left(x-1\right)}=32\)
\(\Rightarrow2^{x+3\left(x-1\right)}=32=2^5\Leftrightarrow x+3\left(x-1\right)=5\)
\(\Rightarrow x+3x-1.3=5\Leftrightarrow1x+3x=4x=5-3=2\)
\(\Rightarrow x=2\div4=\frac{2}{4}\) Vậy \(x=\frac{2}{4}\)
CL
0
KA
1
16 tháng 9 2021
1) \(6x+3=0\Leftrightarrow6x=-3\Leftrightarrow x=-\dfrac{1}{2}\)
2) \(-5x-7=0\Leftrightarrow-5x=7\Leftrightarrow x=-\dfrac{7}{5}\)
3) \(\left(3x-2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=5\end{matrix}\right.\)
4) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
6) \(x^4+8=0\)( vô lý do \(x^4+8\ge8>0\))
Vậy \(S=\varnothing\)
2^x.8^(x-1)=32
=2^x.2^3.(x-1)=25
=2x+3(x-1)=2^5
=x+3(x-1)=5
=x+3x-3=5
=4x-3=5
=>4x=5+3=8
=>x=8:4=2
Vậy x= 2
Giải theo cách lp 8
\(2^x.8^{x-1}=32\)
\(2^x.8^x:8=32\)
\(\left(2.8\right)^x=4\)
\(16^x=4\)
\(\left(2^4\right)^x=2^2\)
\(2^{4x}=2^2\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)