(\(x\) + \(\dfrac{1}{2}\))² = \(\dfrac{1}{16}\)

\(\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- \(\dfrac{3}{4};-\dfrac{1}{4}\)}

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

\(x\)              = (-\(\dfrac{1}{3}\)).(-\(\dfrac{1}{3}\))³

\(x\)             = \(\dfrac{1}{81}\)

Vậy \(x=\dfrac{1}{81}\)

(\(\dfrac{2}{3}\))⁵\(x\) = \(\dfrac{2}{5}\)

       \(x\) = \(\dfrac{2}{5}\) : (\(\dfrac{2}{3}\))⁵ 

       \(x\) = \(\dfrac{243}{80}\)

Vậy \(x\) = \(\dfrac{243}{80}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{75}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{11}{75}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{11}{75}:\frac{1}{2}=\frac{22}{75}\Leftrightarrow\frac{1}{x+2}=\frac{1}{25}\Leftrightarrow x=23\)

BẠN GHI ĐỀ LẠI CÓ DẤU PHẨY ĐI Ạ

5 . y . \(\frac{1}{2}\). x³y(\(\frac{-1}{3}\).x².y)³= \(\frac{5}{2}\)x³y² \(\frac{-1}{27}\) x⁶y³= \(\frac{-5}{54}\)x⁹y⁵

Hệ số \(\frac{-5}{54}\) 

Phần biến : x⁹y⁵

Bậc : 14

Chúc bạn học tốt !!!

51/30

24/65

chưa xong 

27,3 cộng 5,7 = bao nhiêu

fffff

Em nên gõ công thức trực quan để đề bài rõ ràng nhé

3 nhân 2/3 bao nhiêu

ko hiểu...

1/3 . x + 2/5 . ( x-1)

1/3 . x + 2/5 . x  - 2/5

x. ( 1/3 + 2/5 ) - 2/5

x . 11/15 - 2/5

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