tìm x biết:2^(x+1)=32
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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
\(\left(\dfrac{1}{16}\right)^{x+2}=\left(\dfrac{1}{32}\right)^6\)
\(\Leftrightarrow\dfrac{1}{16^{x+2}}=\dfrac{1}{32^6}\)
\(\Leftrightarrow\dfrac{1}{2^{4x+8}}=\dfrac{1}{2^{30}}\)
\(\Leftrightarrow4x+8=30\Leftrightarrow4x=22\Leftrightarrow x=\dfrac{11}{2}\)
(x+2)+(x+4)+(x+6)+...+(x+32)=352(x+2)+(x+4)+(x+6)+...+(x+32)=352
(x+x+...+x)+(2+4+6+...+32)=352(x+x+...+x)+(2+4+6+...+32)=352
16x+272=35216x+272=352
16x=352−272=8016x=352−272=80
x=80:16=5
2x + 1 - 22 = 32
2x + 1 - 4 = 32
2x + 1 = 32 + 4
2x + 1 = 36
2x = 36 - 1
2x = 35
Vậy không có số x thỏa mãn
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\Leftrightarrow\frac{2^x}{2}+5.\frac{2^x}{2^2}=\frac{7}{32}\Leftrightarrow2^x\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\Leftrightarrow2^x=\frac{1}{8}=2^{-3}\)
<=> x=-3
3(x+2)^2+(2x-1)^2-7(x+3)(x-3)=32
3(x^2+4x+4)+(4x^2-4x+1)-7(x^2-9)=32
3x^2+12x+12+4x^2-4x+1-7x^2+81=32
8x+94=32
8x=-62
x=-62÷8
x=-31/4
2 mũ bn z bn
2^( x + 1 ) = 2^5
=> x + 1 = 5
=> x = 5 - 1
=> x = 4
vậy x = 4