\(\dfrac{x}{4}=\dfrac{y}{2}\) 

 ⇒ (\(\dfrac{x}{4}\) )² = \(\dfrac{xy}{4.2}\) = \(\dfrac{72}{8}\) = 9

⇒ \(x^2\) = 4².9

⇒\(x^2\) = 4².3²

⇒ \(x^2\) = 12²

\(\left[{}\begin{matrix}x=-12\\x=12\end{matrix}\right.\)

thay \(x=-12\) vào biểu thức : \(xy\) = 72 ta có :

-12 .\(y\) = 72 => \(y=72:(-12)\) ⇒ \(y=-6\)

thay \(x=12\) vào biểu thức \(xy=72\) ta có :

12\(y\) = 72 ⇒ \(y\) = 72: 12 ⇒ \(y\) = 6

kết luận :

(x;y) =(-12; -6) ; ( 12; 6)

\(\dfrac{x}{4}=\dfrac{y}{2}\) và \(xy=72\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{xy}{4\cdot2}=\dfrac{72}{8}=9\)

\(\left\{{}\begin{matrix}x=9\cdot4=36\\y=9\cdot2=18\end{matrix}\right.\)

Vậy x=36;y=18

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

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

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 !