a.Tìm x biết :(x+1/2)+(x+1/4)+(x+1/8)+(x+1/16)=1
b.Tính:56,24+75,06x4,8-6,15x13,6=.....
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
BK
2
WH
29 tháng 4 2018
Trả lời
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\frac{8+4+2+1}{16}=\frac{23}{16}\)
\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)
\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)
\(\Leftrightarrow4x=\frac{8}{16}\)
\(\Leftrightarrow4x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:4\)
\(\Leftrightarrow x=\frac{1}{8}\)
Vậy x=\(\frac{1}{8}\)
29 tháng 4 2018
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\left(\frac{8+4+2+1}{16}\right)=\frac{23}{16}\)
\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)
\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)
\(\Leftrightarrow4x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:4\)
\(\Leftrightarrow x=\frac{1}{8}\)
TT
3
LH
7 tháng 12 2015
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\text{x}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\)
\(\Leftrightarrow4\text{x}+\frac{15}{16}=1\)
\(\Leftrightarrow4\text{x}=1-\frac{15}{16}\)
\(\Leftrightarrow4\text{x}=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{16}:4=\frac{1}{64}\)
7 tháng 12 2015
(x+1/2)+(x+1/4)+(x+1/8)+(x+16)=1
(x.x.x.x)+(1/2+1/4+1/8+1/16=1
4x+(1/2+1/4+1/8+1/16)=1
4x+(8/16+4/16+2/16+1/16)=1
4x+15/16=1
4x=1-15/16
4x=1/16
x=1/16:4
=>x=1/64
tick cho mk nha bạn
TM
3
DP
30 tháng 6 2017
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{64}\)
ML
30 tháng 6 2017
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}\div4\)
\(x=\frac{1}{64}\)
Vậy ...
T
6
15 tháng 7 2015
ta có:(x+x+x+x)+(1/2+1/4+1/8+1/16)=1 4x+15/16=1 4x=1-15/16 4x=1/16 x=1/16:4=1/64 vậy x=1/64 Mình trả lời trước nhé
N
15 tháng 7 2015
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\)
\(A=1-\frac{1}{2^4}\)
\(A=\frac{15}{16}\)
Thay A vào đẳng thức ta có
\(4x+\frac{15}{16}=1\)
\(4x=1-\frac{15}{16}\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}\div4\)
\(x=\frac{1}{64}\)
TT
0
10 tháng 1
Bài 1:
a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)
\(P=\left(\dfrac{x+1}{3x^2+3x}+\dfrac{1-2x}{6x^2-3x}-1\right):\dfrac{1-x}{2x}\)
\(=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{-\left(x-1\right)}\)
\(=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{-2x}{x-1}\)
\(=\left(-1\right)\cdot\dfrac{-2x}{x-1}=\dfrac{2x}{x-1}\)
b: Để P nguyên thì \(2x⋮x-1\)
=>\(2x-2+2⋮x-1\)
=>\(2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{2;3\right\}\)
c: P<1
=>P-1<0
=>\(\dfrac{2x}{x-1}-1< 0\)
=>\(\dfrac{2x-x+1}{x-1}< 0\)
=>\(\dfrac{x+1}{x-1}< 0\)
=>-1<x<1
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}-1< x< 1\\x\ne0\end{matrix}\right.\)
TK
2
MH
31 tháng 12 2015
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)
\(\Rightarrow4\times x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Rightarrow4\times x+\frac{8+4+2+1}{16}=1\)
\(\Rightarrow4\times x+\frac{15}{16}=1\)
\(\Rightarrow4\times x=1-\frac{15}{16}=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{16}:4=\frac{1}{16}\times\frac{1}{4}\)
\(\Rightarrow x=\frac{1}{64}\).
15 tháng 12 2021
1) Áp dụng HTL:
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{4^2}+\dfrac{1}{\left(4\sqrt{2}\right)^2}=\dfrac{3}{32}\Rightarrow AH=\dfrac{4\sqrt{6}}{3}\left(cm\right)\)
Áp dụng đ/lý Pytago:
\(BC^2=AB^2+AC^2\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+\left(4\sqrt{2}\right)^2}=4\sqrt{3}\left(cm\right)\)
Bài 2:
a) \(pt\Leftrightarrow\sqrt{\left(2x+1\right)^2}=3\Leftrightarrow\left|2x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b) \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=2\sqrt{x}.\dfrac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+2\)
8 tháng 3 2016
(x+1/2)+(x+1/4)+(x+1/8)+(x+1/16)=1
4x+(1/2+1/4+1/8+1/16)=1
4x+15/16=1
4x=1-15/16
4x=1/16
x=1/64