\(\left(\dfrac{1}{2}\right)^{12}=\left(\dfrac{1}{8}\right)^3\\ \left(\dfrac{1}{3}\right)^9=\left(\dfrac{1}{27}\right)^3\\ Ta\text{ }có:\dfrac{1}{8}>\dfrac{1}{27}\\ Vậy:\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)

\(\left(\dfrac{1}{2}\right)^{12}=\dfrac{1}{4096};\left(\dfrac{1}{3}\right)^9=\dfrac{1}{19683}\\ \Rightarrow\dfrac{1}{4096}>\dfrac{1}{19683}\\ \Rightarrow\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)

cho : b² = ac

đem vào bên trái CMR , tính ra bên phải

\(Theo\text{ }bài\text{ }ra:2a=3b=4c\\ \Rightarrow\dfrac{2a}{12}=\dfrac{3b}{12}=\dfrac{4c}{12}\\ \Rightarrow\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}\\ \RightarrowĐặt\text{ }\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=k\\ \Rightarrow\left\{{}\begin{matrix}a=6k\\b=4k\\c=3k\end{matrix}\right.\\ Khi\text{ }đó\dfrac{a-b+c}{a+2b-c}=\dfrac{6k-4k+3k}{6k+8k-3k}=\dfrac{5k}{11}=\dfrac{5}{11}\\ Vậy:A=\dfrac{5}{11}.\)

`a, 16/x = x /25`

`<=> 16 . 25 = x^2`

`<=> 400 = x^2`

`<=> x = +-20`.

`b, x/-2 = -8/x`

`<=> x^2 = (-2).(-8)`

`<=> x^2 = 16`

`<=> x = +-4`.`

c, -4/x = x/-49`

`x^2 = (-4).(-49)`

`x^2 = 196`

`x = +-14.`

`d, -x/3 = 27/-x`

`<=> (-x)^2 = 81`

`<=> x^2 = 81`

`<=> x = +-9`

Giúp e làm bài với ạ!! E cảm ơn!

...

          B  = 1 + 2² + 2⁴ +.....+ 2¹⁰⁰

      2²B  =       2² + 2⁴ +.....+ 2¹⁰⁰ + 2¹⁰²

2²B - B  =  2¹⁰² - 1

        3B = (2¹⁰² -1) 

          B = (2¹⁰² - 1) : 3