\(3^2x2^5x\left(\frac{2}{3}\right)^3\)

\(=\frac{256}{3}\)

đúng nhé

tk nữa

\(3^2.2^5.\left(\frac{2}{3}\right)^3\)

\(=9.32.\frac{8}{27}\)

\(=\frac{256}{3}\)

3² x 2⁵ x ( 2/3 ) ³

= 256/3

= 1/2 . 2/3 .... 2014/2015 . 2015/2016

= 1/2016

1/2016

A = ( 1 - 1/2 ) ( 1 - 1/3 ) ... ( 1 - 1/2015 ) ( 1 - 1/2016 )

A = 1/2 . 2/3 ... 2014/2015 . 2015/2016

A = 1 . 2 . 3 ... 2014 . 2015/2 . 3 ... 2014 . 2015 . 2016

A = 1/2016

Nếu đúng thì k cho mình nha

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)

\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)

\(\Leftrightarrow\frac{38}{x+3}=16\)

\(\Leftrightarrow x+3=2,375\)

\(\Leftrightarrow x=-0,625\)

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)

\(\Leftrightarrow\frac{38}{x+3}=14\)

\(\Leftrightarrow\left(x+3\right)14=38\)

\(\Leftrightarrow14x+42=38\)

\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)

Vậy \(x=-\frac{2}{7}\)

TRẻ Trâu 

\(\left(x+\frac{1}{2}\right)\left(x-\frac{3}{4}\right)=0\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

\(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)

\(x+\frac{1}{2}=0\)hoặc \(x-\frac{3}{4}=0\)

\(\Leftrightarrow x=-\frac{1}{2}\)hoặc \(x=\frac{3}{4}\)

\(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)

\(< =>\hept{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\)

TH 1: \(x+\frac{1}{2}=0\)

\(x=0-\frac{1}{2}\)

\(x=-\frac{1}{2}\) (1)

TH 2: \(x-\frac{3}{4}=0\)

\(x=\frac{3}{4}+0\)

\(x=\frac{3}{4}\) (2)

Từ (1) và (2) => x thuộc tập hợp \(\left(-\frac{1}{2};\frac{3}{4}\right)\)

chúc bạn học tốt nha

\(-4\frac{1}{2}\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

<=>  \(-1,5\le x\le\frac{11}{18}\)

đến đây tự làm

mk k biết điều kiện của x  nên giúp đến đó

\(\Leftrightarrow2.\left(\frac{-1}{2}\right).\left(\frac{2}{3}\right)^2-3\left(-\frac{1}{3}\right)^2.\frac{2}{9}:x=3.\left(-\frac{1}{2}\right)-\frac{2}{3}\)

\(\Leftrightarrow-\frac{4}{9}-\frac{1}{3}.\frac{2}{9}:x=-\frac{3}{2}-\frac{2}{3}\)

\(\Leftrightarrow-\frac{4}{6}-\frac{2}{27}:x=-\frac{13}{6}\)

\(\Leftrightarrow\frac{2}{27}:x=-\frac{4}{9}:\frac{-13}{6}\)

\(\Leftrightarrow\frac{2}{27}:x=\frac{31}{18}\)

\(\Leftrightarrow x=\frac{2}{27}:\frac{31}{18}\)

\(\Rightarrow x=\frac{4}{93}\)

Vậy \(x=\frac{4}{93}\)