A m x y n

Tên các góc vuông là :

\(\widehat{xAm};\widehat{mAy};\widehat{nAy};\widehat{nAx}\)

~~Học tốt~~

thank...

có : \(y\inℕ\)

\(\Rightarrow m-3⋮m+2\)

\(\Rightarrow m+2-5⋮m+2\)

có :\(m+2⋮m+2\)       \(\Rightarrow-5⋮m+2\)

\(\Rightarrow m+2\inƯ_{\left(5\right)}\)

    Vậy để \(y\inℕ\) thì  \(m=\left\{-7;-3;-1;3\right\}\)

Mn ơiii giúp mk ik mà

Mn ơiii giúp mk với

Ta có :

\(x^2=\frac{24}{25}\)

\(\Rightarrow x=\pm\sqrt{\frac{24}{25}}\)

\(\Rightarrow x\in\left\{\sqrt{\frac{-24}{25}};\sqrt{\frac{24}{25}}\right\}\)

Cậu ơi tớ chưa học tới Căn bậc ạ.

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

Dùng KT \(\left|x\right|\ge0\)\(\forall\)\(x\)

BG :

Ta có : \(\left|x-\frac{1}{2}\right|\ge0\)\(\forall\)\(x\)

\(\Rightarrow\)\(\left|x-\frac{1}{2}\right|+\frac{3}{4}\ge0+\frac{3}{4}\)\(\forall\)\(x\)

\(\Rightarrow\)\(\left|x-\frac{1}{2}\right|+\frac{3}{4}\ge\frac{3}{4}\)\(\forall\)\(x\)

Hay \(B\ge\frac{3}{4}\)\(\forall\)\(x\)

Dấu "=" xảy ra khi :

 \(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=0\)

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

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy GTNN của \(B=\frac{3}{4}\)đạt được khi \(x=\frac{1}{2}\)

\(A=\left|x+\frac{3}{2}\right|\ge0\)

\(MinA=0\Rightarrow\left|x+\frac{3}{2}\right|=0\Rightarrow x=\frac{-3}{2}\)

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

\(B\ge\frac{3}{4}\)do\(\left|x-\frac{1}{2}\right|\ge0\)

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