2x + 32 = 32

2x = 0

x = 0

Chọn A

A nha

\(ĐXKĐ:x\ne2\)

\(\dfrac{2-x}{32}=\dfrac{2}{2-x}\\ \Leftrightarrow\left(2-x\right)^2=64\\ \Rightarrow\left[{}\begin{matrix}2-x=-8\\2-x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(\dfrac{2x-x}{32}=\dfrac{2}{2-x}\\ \Rightarrow\dfrac{x}{32}=\dfrac{2}{2-x}\\ \Rightarrow x\left(2-x\right)=64\\ \Rightarrow-x^2+2x-64=0\\ \Rightarrow x^2-2x+64=0\\ \Rightarrow\left(x^2-2x+1\right)+63=0\\ \Rightarrow\left(x-1\right)^2+63=0\)

Vì \(\left(x-1\right)^2\ge0;63>0\Rightarrow\left(x-1\right)^2+63>0\)

Vậy không có x thỏa mãn đề bài

(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

a) x = 3.     

b) x Î {-3; 3}. 

c) x = 5.

d) x Î {-7; 7). 

x  ∈ {-7;7}

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

<=> x=-3