c) Ta có: \(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\forall x\in Q\\\left|2,5-x\right|\ge0\forall x\in Q\end{matrix}\right.\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\forall x\in Q\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left|x-1,5\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

Vậy \(x=\left\{{}\begin{matrix}1,5\\2,5\end{matrix}\right.\).

e) \(\left(x-2\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}x-2=\sqrt{1}\\x-2=-\sqrt{1}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\).

Mấy câu kia dễ rồi.

sửa lại ý c của N.Anh

Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(\left|x-1,5\right|+\left|2,5-x\right|\ge\left|x-1,5+2,5-x\right|=1\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge1>0\)

mà theo đề thì \(\left|x-1,5\right|+\left|2,5-x\right|=0\)

\(\Rightarrow\) k có gt \(x\) nào tm yêu cầu đề bài

a) \(P=\frac{3x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(P=\frac{3\left(x-9\right)}{\left(x-3\right)\left(x-2\right)}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(P=\frac{3}{x-2}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

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

\(P=\frac{9-3x-9+x^2-2x^2+4x-x+2}{\left(x-2\right)\left(3-x\right)}\)

\(P=\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}\) (*)

b) Thay \(x=-\frac{1}{2}\) vào (*) ta có:

\(P=\frac{2-\left(-\frac{1}{2}\right)^2}{\left[\left(-\frac{1}{2}\right)-2\right]\left[3-\left(-\frac{1}{2}\right)\right]}=\frac{2-\frac{1}{4}}{-\frac{5}{2}.\frac{7}{2}}=-\frac{\frac{7}{4}}{\frac{5}{2}.\frac{7}{2}}=-\frac{7}{35}=-\frac{1}{5}\)

c) \(\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}< 0\)

\(\Leftrightarrow2-x^2< 0\)

\(\Leftrightarrow-x^2< -2\)

\(\Leftrightarrow x^2>2\)

\(\Leftrightarrow\hept{\begin{cases}x< -\sqrt{2}\\-\sqrt{2}< x< \sqrt{2}\\x>2\end{cases}}\)

Vậy: ...

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-3=0\\x-\frac{1}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)

Vậy \(x=3\) hoặc \(x=\frac{1}{2}\)

\(b)\) \(x^2-2x=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

Vậy \(x=0\) hoặc \(x=2\)

\(c)\) \(\left(3x-1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-1=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=1\\x^2=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x\in\left\{\varnothing\right\}\end{cases}}}\)

Vậy \(x=\frac{1}{3}\)

\(d)\) \(\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy \(x=-1\) hoặc \(x=2\)

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