\(|\) x-1/\(|\) , \(|\)y-3/ \(|\) ≥ 0 với mọi x,y

 x-1=0

 x    =0+1=1

y-3 =0

y    = 0+1=3

Vậy x=1, y=3

Xét tg vuông ABN và tg vuông ACN có

AB=AC (gt)

\(\widehat{A}\) chung

=> tg AMN = tg ACM (Hai tg vuông có cạnh huyền và góc nhọ tương ứng bằng nhau) \(\Rightarrow\widehat{ACM}=\widehat{ABN}\)

( \(\dfrac{1}{9}+\dfrac{2}{3}\))² - \(\dfrac{5}{3}:5\)

= (\(\dfrac{7}{9}\))² - \(\dfrac{1}{3}\)

= \(\dfrac{49}{81}\) - \(\dfrac{1}{3}\)

= \(\dfrac{22}{81}\)

0.27160493827 nha bạn!

\(\dfrac{x}{2}=\dfrac{y}{3}\)  => x = \(\dfrac{2}{3}\) .y 

\(\dfrac{y}{4}=\dfrac{z}{5}\) => z = \(\dfrac{5}{4}\).y

thay x = \(\dfrac{2}{3}\).y và z = \(\dfrac{5}{4}\).y vào biểu thức x + 2y - z = 28 ta có:

\(\dfrac{2}{3}\)y + 2y - \(\dfrac{5}{4}\) y = 28

y.( \(\dfrac{2}{3}+2-\dfrac{5}{4}\)) = 28

y. \(\dfrac{17}{12}\) = 28

y = 28 : \(\dfrac{17}{12}\)

y = \(\dfrac{336}{17}\)

x = \(\dfrac{2}{3}\). \(\dfrac{336}{17}\)

x = \(\dfrac{224}{17}\)

z = \(\dfrac{5}{4}\) . \(\dfrac{336}{17}\)

z = \(\dfrac{420}{17}\)

vậy (x,y,z) = (\(\dfrac{224}{17}\); \(\dfrac{336}{17}\); \(\dfrac{420}{17}\))

\(\sqrt{\left(5x+1\right)^2}=\dfrac{6}{7}\)

\(\Leftrightarrow\left|5x+1\right|=\dfrac{6}{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\)