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

bạn học lớp mấy vậy

lâu lắm rồi mình mới onl lại

2^x + 1 - 2² = 32

2^x + 1 - 4 = 32

2^x + 1 = 32 + 4

2^x + 1 = 36

2^x = 36 - 1

2^x = 35

Vậy không có số x thỏa mãn

x không tồn tại vì 2 ^ x +1 là số lẻ nên không thể bằng 32

2^x+1 - 2^x = 32

2^x. 2 - 2^x = 32

2^x. (2-1) = 32

2^x = 32

x = 5

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\Leftrightarrow\frac{2^x}{2}+5.\frac{2^x}{2^2}=\frac{7}{32}\Leftrightarrow2^x\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\Leftrightarrow2^x=\frac{1}{8}=2^{-3}\)

<=> 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

\(2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=5-1\)

\(\Rightarrow x=4\)

Vậy  \(x=4\)