\(D=4^2+4^3+...+4^{25}\)

\(\Rightarrow4D=4^3+4^4+...+4^{26}\)

\(\Rightarrow3D=4D-D=4^3+4^4+...+4^{26}-4^2-4^3-...-4^{25}=4^{26}-4^2\)

\(\Rightarrow D=\dfrac{4^{26}-4^2}{3}\)

đáp án bàng bao nhiêu vậy

\(\frac{2}{9}\div\frac{2}{9}\times\frac{1}{2}\)

\(=\frac{2}{9}\times\frac{9}{2}\times\frac{1}{2}\)

\(=\frac{2\times9\times1}{9\times2\times2}\)

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

TL

\(\frac{2}{9}:\frac{2}{9}x\frac{1}{2}\)

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

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

1/2* x+2/3=9/2

1/2 * x = 9/2 - 2/3 

1/2 * x= 23/6

x= 23/6 : 1/2

x= 23/6 x 2= 23/3

___

1/2*x-1/3=2/3

1/2*x = 2/3 + 1/3

1/2 * x= 1

x= 1: 1/2 

x= 2

____

1/4+3/4:x=3

3/4 : x = 3 - 1/4

3/4 : x= 11/4

x= 11/4 : 3/4

x= 11/3

\(\dfrac{1}{2}\)\(\times\)\(x\) + \(\dfrac{2}{3}\) = \(\dfrac{9}{2}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)        = \(\dfrac{9}{2}\) - \(\dfrac{2}{3}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)       = \(\dfrac{23}{6}\)

      \(x\)       = \(\dfrac{23}{6}\):\(\dfrac{1}{2}\)

      \(x\)      = \(\dfrac{23}{3}\) 

\(\dfrac{1}{2}\)\(\times\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\) 

\(\dfrac{1}{2}\)\(\times\)\(x\)       = \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)

\(\dfrac{1}{2}\times\)\(x\)      =  1

     \(x\)       = 1 : \(\dfrac{1}{2}\)

   \(x\)         = 2

\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\): \(x\) = 3

          \(\dfrac{3}{4}\): \(x\) = 3 - \(\dfrac{1}{4}\) 

          \(\dfrac{3}{4}\):\(x\) = \(\dfrac{11}{4}\)

              \(x\) = \(\dfrac{3}{4}\): \(\dfrac{11}{4}\)

             \(x\) = \(\dfrac{3}{11}\)

y x 2/3 + 1/3 = 8/9 : 2

Y x 2/3 +1/3 = 4/9

Y x 1/3 = 4/9 - 1/3

y x 1/3 = 1/9

Y = 1/9 : 1/3

y = 1/3

y x 2/3 + 1/3 = 8/9 : 2

y x 2/3 + 1/3 = 4/9

                y   = (4/9 - 1/3) : 2/3

                y   = 1/6

( dùng thứ lại biểu thức cộng trừ trước nhân chia sau)

a: x>=9

=>x-9>=0

=>A=x-9+7=x-2

b: x<0

=>-3x>0

=>\(B=-3x-8x^2+x-2=-8x^2-2x-2\)

c: TH1: x>=1

=>x-1>=0

C=3x+|x-1|-2

=3x+x-1-2

=4x-3

TH2: x<1

=>x-1<0

C=3x+|x-1|-2

=3x-2-x+1

=2x-1

d: x<0

=>-x>0

\(D=\frac{x^2-\left|2x\right|+1}{\left(x+1\right)\cdot\left|-x\right|}\)

\(=\frac{x^2+2x+1}{\left(x+1\right)\cdot\left(-x\right)}\)

\(=\frac{\left(x+1\right)^2}{-x\left(x+1\right)}=\frac{x+1}{-x}\)

a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.

b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.

Lời giải:

a.

$|4x-1|-|3x-\frac{1}{2}|=0$

$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)

b. Nếu $x\geq 1$ thì:

$|x-1|-2x=\frac{1}{2}$

$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$

$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)

Nếu $x< 1$ thì:

$1-x-2x=\frac{1}{2}$

$\Leftrightarrow x=\frac{1}{6}$ (tm)

\(\left|x+4\right|=21\)

\(\Rightarrow\orbr{\begin{cases}x+4=21\\x+4=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=17\\x=-25\end{cases}}}\)

Vậy ...

mình có thể làm bạn với bạn

\(a,\Rightarrow\dfrac{\left(-3\right)^x}{\left(-3\right)^4}=\left(-3\right)^3\\ \Rightarrow\left(-3\right)^{x-4}=\left(-3\right)^3\\ \Rightarrow x-4=3\Rightarrow x=7\\ b,Sửa:\left(x-\dfrac{1}{2}\right)^2=25\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=5\\x-\dfrac{1}{2}=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{5}\\x=-\dfrac{9}{5}\end{matrix}\right.\)