\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\)

\(Đặt:A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\)

Ta có:

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\)

\(\dfrac{A}{2}=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{91}}\)

\(\dfrac{A}{2}-A=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{91}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\right)\)

\(-\dfrac{1}{2}A=\dfrac{1}{2^{91}}-\dfrac{1}{2}\)

\(A=\dfrac{\left(\dfrac{1}{2^{91}}-\dfrac{1}{2}\right)}{-\dfrac{1}{2}}\)

\(A=-2\left(\dfrac{1}{2^{91}}-\dfrac{2^{90}}{2^{91}}\right)\)

\(A=\dfrac{-2\left(1-2^{90}\right)}{2^{91}}\)

\(A=\dfrac{-2-\left[-\left(2^{91}\right)\right]}{2^{91}}\)

\(A=\dfrac{-2+2^{91}}{2^{91}}\)

\(A=-\dfrac{2}{2^{91}}+\dfrac{2^{91}}{2^{91}}\)

\(A=\dfrac{1}{2^{90}}-1\)

2x-0,5=x+\(\dfrac{1}{4}\)

<=> 2x-x=\(\dfrac{1}{4}+0,5\)

<=> x= 0,75

a) $|6,5 - x| = 35$

$6,5 - x= 35$ hoặc $6,5 - x = -35$

$x = -28,5$ hoặc $x = 41,5$.

b) $\dfrac13 + \dfrac23$ : $x = -1 + \left|-\dfrac23\right|$

$\dfrac13 + \dfrac23$ : $x = -1 + \dfrac23$

$\dfrac23$ : $x = -\dfrac23$

$x = -1$.

\(\dfrac{4}{9}-\dfrac{2}{3}.x=\dfrac{1}{3}\\ \dfrac{2}{3}.x=\dfrac{4}{9}-\dfrac{1}{3}\\ \dfrac{2}{3}.x=\dfrac{1}{9}\\ x=\dfrac{1}{9}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{1}{6}\)

1/6

\(=\dfrac{1}{3}.\dfrac{3}{5}-\left(\dfrac{5}{6}\right)^2\\ =\dfrac{1}{5}-\dfrac{25}{36}\\ =-\dfrac{89}{180}\)

chịu

a) xét ΔABH và ΔACH, ta có :

AB = AC (giả thiết)

\(\widehat{ABC}=\widehat{ACB}\)  (vì AB = AC => đó là tam giác cân, mà tam giác cân thì có 2 góc ở đáy bằng nhau)

AH là cạnh chung

ð ΔABH = ΔACH (c.c.c)

b) vì ΔABH = ΔACH, nên :

=> HB = HC (2 cạnh tương ứng)

c) hơi khó nha !

( \(\dfrac{1}{125}\) - \(x^3\) ) ( \(x^2\) + 2².6⁶) = 0

\(\left[{}\begin{matrix}\dfrac{1}{125}-x^3=0\\x^2-2^2.6^6=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x^3=\dfrac{1}{125}\\x^2=2^2(6^3)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x^2=(2.6^3)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=432^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=432\\x=-432\end{matrix}\right.\)

\(x\) ϵ { -432; \(\dfrac{1}{5}\); 432; }

a) xét ΔOCB và ΔODA, ta có :

OA = OB (giả thiết)

\(\widehat{O}\) là góc chung

AC = BD (giả thiết)

⇒ ΔOCB = ΔODA (c.g.c)

⇒ AC = BD (2 cạnh tương ứng)

b) xét ΔEAC và ΔEBD, ta có : 

AD = BC (câu a)

\(\widehat{AEC}=\widehat{BED}\) (vì là 2 góc đối đỉnh) 

AC = BD (giả thiết)

⇒ ΔEAC = ΔEBD (C.G.C)

c) xét ΔOAE và ΔOBE, ta có :

OA = OB (giả thiết)

AE = BE [vì ΔEAC = ΔEBD (2 cạnh tương ứng)]

OE là cạnh chung

⇒ ΔOAE = ΔOBE (c.c.c)

⇒ \(\widehat{AOE}=\widehat{BOE}\) (2 góc tương ứng)