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

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\(\left|x-\frac{1}{2}\right|\ge0\)

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

\(\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)

\(=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)

\(=\frac{-9}{16}\)

Mn ơiii giúp mk với ạ mk sẽ k cho mn

Mn ơiii giúp mk với ạ

Ta có: 5x2.3xy2 = (5.3).(x2.x).y2 = 15x3y2

Phần hệ số: 15

\(\frac{13}{x-5}\left(ĐKXĐ:x\ne5\right)\)

Để: \(\frac{13}{x-5}\inℤ\Rightarrow13⋮x-5\)

\(\Rightarrow x-5\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\Rightarrow x\in\left\{6;4;18;-8\right\}\)

\(\frac{x+3}{x-2}\)

Để \(\frac{x+3}{x-2}\inℤ\Rightarrow x+3⋮x-2\)mà\(x-2⋮x-2\)

\(\Rightarrow x+3-\left(x-2\right)⋮x-2\)

\(\Rightarrow x+3-x+2⋮x-2\)

Mà: \(x-2\inℤ\Rightarrow x-2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\Rightarrow x\in\left\{3;1;7;-3\right\}\)

So sánh \(\frac{-19}{49}\)và\(\frac{-23}{27}\)

Quy đồng lên ta được: \(\frac{-19}{49}=\frac{\left(-19\right).47}{49}=\frac{-893}{2303}\)và\(\frac{-23}{47}=\frac{\left(-23\right).49}{47.49}=\frac{-1127}{2303}\)

Mà: \(\frac{-893}{2303}>\frac{-1127}{2303}\)\(\Rightarrow\frac{-19}{49}>\frac{-23}{47}\)

em làm xong từ hôm qua r ạ

thôi thôi thôi thôi thôi thôi thôiiiiiiiiiiiiiiiiii

nghỉ đi bạn

\(\frac{1}{3}-\frac{3}{4}-\left(\frac{-3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)

\(=\left(\frac{24}{72}-\frac{54}{72}+\frac{1}{72}-\frac{16}{72}-\frac{2}{72}\right)+\left(\frac{1}{15}+\frac{9}{15}\right)\)

\(=\frac{-47}{72}+\frac{2}{3}\)

\(=\frac{1}{72}\)