\(a,\)\(\frac{1}{2}-\frac{3}{4}.\frac{-6}{5}\)

\(=\frac{1}{2}-\frac{-9}{10}\)

\(=\frac{1}{2}+\frac{9}{10}\)

\(=\frac{7}{5}\)

\(b,\frac{3}{8}+2^2-\frac{3}{8}\)

\(=\left(\frac{3}{8}-\frac{3}{8}\right)+2^2\)

\(=0+4\)

\(=4\)

a) \(\frac{1}{2}-\frac{3}{4}.\left(-\frac{6}{5}\right)\)\(=\frac{1}{2}-\left(-\frac{9}{10}\right)\)\(=\frac{1}{2}+\frac{9}{10}=\frac{5}{10}+\frac{9}{10}=\frac{14}{10}=\frac{7}{5}\)

b) \(\frac{3}{8}+2^2-\frac{3}{8}=\frac{3}{8}+4-\frac{3}{8}=\frac{3}{8}-\frac{3}{8}+4=0+4=4\) 

Hoặc \(\frac{3}{8}+2^2-\frac{3}{8}=\frac{3}{8}+4-\frac{3}{8}=\frac{3}{8}+\frac{32}{8}-\frac{3}{8}=\frac{35}{8}-\frac{3}{8}=\frac{32}{8}=4\)

Áp dụng ĐL Pi ta go trong

tam giác vuông OAP có: AP² = OA² - OP²

Trong tam giác vuông OAN có: AN² = OA² - ON²

Tương tự, với các tam giác vuông OBP; OBM; OCM; OCN 

Ta có: AN² + BP² + CM² = (OA² - ON²) + (OB² - OP²) + (OC² - OM²)  = (OA² + OB² + OC²) - (ON² + OP² + OM²) 

AP² + BM² + CN² = (OA² - OP²) + (OB² - OM²) + (OC² - ON²) = (OA² + OB² + OC²) - (ON² + OP² + OM²) 

=>  AN² + BP² + CM²  = AP² + BM² + CN²

a) \(1-2x< 7\)

\(\Rightarrow2x>-6\)

\(\Rightarrow x>-3\)

Vậy \(x>-3\)

b) \(\left(x-1\right)\left(x-2\right)>0\)

\(\Rightarrow\hept{\begin{cases}x-1>0\\x-2>0\end{cases}}\)

hoặc \(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x>1\\x>2\end{cases}}\)

hoặc \(\hept{\begin{cases}x< 1\\x< 2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>2\\x< 1\end{cases}}\)

Vậy \(x>2\)hoặc \(x< 1\)

c) \(\left(x-2\right)^2\left(x-1\right)\left(x-4\right)< 0\left(1\right)\)

Vì \(\left(x-2\right)^2\ge0\forall x\)nên từ \(\left(1\right)\): \(\Rightarrow\left(x-2\right)^2>0\)

\(\Rightarrow\orbr{\begin{cases}x-2>0\\x-2< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>2\\x< 2\end{cases}}\)

Với : 

• \(x>2\)

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

Nếu \(-2< x-4< 0\)thì \(2< x< 4\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)(Thỏa mãn)

Nếu \(x-4\ge0\)thì \(x\ge4\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)\ge0\)(Không thỏa mãn)

• \(x< 2\)

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

Nếu \(0< x+1< 3\)thì \(-1< x< 2\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)(Thỏa mãn)

Nếu \(x+1\le0\)thì \(x\le-1\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)\ge0\)(Không thỏa mãn)

Vậy \(2< x< 4\)hoặc \(-1< x< 2\)

d) \(\frac{5}{x}< 1\)

\(\Rightarrow5< x\)

Vậy \(x>5\)