Tìm x: \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
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\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(4x+\frac{15}{16}=\frac{23}{16}\)
\(4x=\frac{1}{2}\)
\(x=\frac{1}{8}\)
Vậy \(x=\frac{1}{8}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Rightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Rightarrow5x+\frac{15}{32}=\frac{23}{16}\)
\(\Rightarrow5x=\frac{23}{16}-\frac{15}{32}\)
\(\Rightarrow5x=\frac{31}{32}\)
\(\Rightarrow x=\frac{31}{32}.\frac{1}{5}=\frac{31}{160}\)
câu này ở trong Violympic nên mình nói luôn đáp án là 1/8
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Rightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\frac{15}{16}=1\)
\(\Rightarrow4x=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{64}\)
=>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 \(x=\frac{1}{64}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow x+x+x+x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\)
\(\Leftrightarrow4x+\frac{15}{16}=1\Leftrightarrow4x=\frac{1}{16}\Leftrightarrow x=\frac{1}{16}:4=\frac{1}{64}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)
\(\Rightarrow4x+\frac{15}{16}=1\)
\(\Rightarrow4x=1-\frac{15}{16}=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{16}:4=\frac{1}{16}.\frac{1}{4}=\frac{1}{64}\)
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
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