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

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

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

\(\Leftrightarrow4x+\left(\frac{8+4+2+1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)

\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)

\(\Leftrightarrow4x=\frac{1}{2}\)

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

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

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

\(\Leftrightarrow4\text{x}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\)

\(\Leftrightarrow4\text{x}+\frac{15}{16}=1\)

\(\Leftrightarrow4\text{x}=1-\frac{15}{16}\)

\(\Leftrightarrow4\text{x}=\frac{1}{16}\)

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

(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

(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

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

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

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

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

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

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

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

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

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

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

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

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

\(x=\frac{1}{16}\div4\)

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

Vậy ...

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

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

=>4x+15/16=1

=>4x=1:15/16

=>4x=1 x 16/15

=>4x=1/16

=>x=1/16:4

=>x=1/64

Vậy x=1/64

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

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

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

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

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

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

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

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

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

\(x=\frac{1}{16}\div4\)

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

cũng như bạn

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: \(\left\{{}\begin{matrix}-1< x< 1\\x\ne0\end{matrix}\right.\)

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

x= 1/64

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

\(\Rightarrow4\times x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(\Rightarrow4\times x+\frac{8+4+2+1}{16}=1\)

\(\Rightarrow4\times x+\frac{15}{16}=1\)

\(\Rightarrow4\times x=1-\frac{15}{16}=\frac{1}{16}\)

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

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

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) \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(=2\sqrt{x}.\dfrac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+2\)

(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