\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(4x+\frac{15}{16}=\frac{23}{16}\)

\(4x=\frac{1}{2}\)

\(x=\frac{1}{8}\)

             Vậy \(x=\frac{1}{8}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Rightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Rightarrow5x+\frac{15}{32}=\frac{23}{16}\)

\(\Rightarrow5x=\frac{23}{16}-\frac{15}{32}\)

\(\Rightarrow5x=\frac{31}{32}\)

\(\Rightarrow x=\frac{31}{32}.\frac{1}{5}=\frac{31}{160}\)

Có 4x thôi Nguyễn Quang Trung

=>4x+(1/2+1/4+1/8+1/16)=1

<=>4x+15/16=1

=>4x=1/16

=>x=1/16:4=1/64

vậy x=1/64

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Rightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(\Rightarrow4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)

\(\Rightarrow4x+\left(1-\frac{1}{16}\right)=1\)

\(\Rightarrow4x+\frac{15}{16}=1\)

\(\Rightarrow4x=1-\frac{15}{16}\)

\(\Rightarrow x=\frac{1}{16}:4\)

\(\Rightarrow x=\frac{1}{64}\)

vậy \(x=\frac{1}{64}\)

truoc tien quy dong roi tinh hoac so sanh voi 1/2 kich nhe

câu này ở trong Violympic nên mình nói luôn đáp án là 1/8

x *4+1/2+1/4+1/8+1/16=1.

từ đó ra

x. 4 +15/16 = 1

x.4 = 1 - 15/16 = 1/16

x = 1/16 : 4 = 1/64

\(x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)

\(4\cdot x+\frac{1+2+4+8}{16}=1\)

\(4\cdot x+\frac{15}{16}=1\)

Vậy 4*x = 1 - 15/16 = 1/16

Nên x = \(\frac{1}{16}:4=\frac{1}{64}\)

= 1 nhé mk nhầm

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Rightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(\Rightarrow4x+\frac{15}{16}=1\)

\(\Rightarrow4x=\frac{1}{16}\)

\(\Rightarrow x=\frac{1}{64}\)

bệnh lười tái phát 

ko chép lại đề

\(x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)

\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(4x+\frac{15}{16}=1\)

\(4x=\frac{1}{16}\)

\(x=\frac{1}{64}\)

Mong các bạn và thầy cô giải giùm ạ!

Đặt \(t=\left(x+\frac{1}{x}\right)^2\)\(\Rightarrow\)\(x^2+\frac{1}{x^2}=t-2\)điều kiện t>=0,x # 0

Phương trình trở thành

8t +4(t-2)² - 4(t-2)²t =(x+4)²

8t + 4t² - 16t + 16 -4t³ + 16t² - 16t=(x+4)²

-4t³ + 20t² -24t=x² +8x

-4t(t² -5t +6)=x(x+8)

-4t(t-2)(t-3)=x(x+8)

Mình chỉ giúp dược tới đó

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow x+x+x+x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\)

\(\Leftrightarrow4x+\frac{15}{16}=1\Leftrightarrow4x=\frac{1}{16}\Leftrightarrow x=\frac{1}{16}:4=\frac{1}{64}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)

\(\Rightarrow4x+\frac{15}{16}=1\)

\(\Rightarrow4x=1-\frac{15}{16}=\frac{1}{16}\)

\(\Rightarrow x=\frac{1}{16}:4=\frac{1}{16}.\frac{1}{4}=\frac{1}{64}\)

 bài nay mình bít cách làm nhưng hk bít đúng hông. cảm ơn các cậu nhiều.