câu 2 sửa lại là: I have bought a laptop last week but I haven't used it.

1. Last year our club donated books for children in rural areas.
2. I have bought a laptop but I haven't used it.

3. Children sent thank-you cards to us a week ago.
4. We started to play the game half an hour ago.

5. I taught two children in grade 2 last summer.
6. I have read my English book but I haven't done my English homework this morning.

7. Last spring we helped the elderly in the nursing home.

8. We helped people in flooded areas last year.

Ta có:

\(f\left(x\right)=5x^2-12x+1=5\left(x^2-\dfrac{12}{5}x+\dfrac{1}{5}\right)=5\left[\left(x^2-2.\dfrac{6}{5}x+\dfrac{36}{25}\right)-\dfrac{36}{25}+\dfrac{1}{5}\right]=5\left[\left(x-\dfrac{6}{5}\right)^2-\dfrac{31}{25}\right]=5\left(x-\dfrac{6}{5}\right)^2-\dfrac{31}{5}\ge-\dfrac{31}{5}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{6}{5}\)

Vậy GTNN của f(x) là \(-\dfrac{31}{5}\) khi \(x=\dfrac{6}{5}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

\(A=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}\)

\(=\dfrac{x-1}{2}\)

b) Yêu cầu bài toán: \(\dfrac{x-1}{2}=-\dfrac{1}{2}\Leftrightarrow x-1=-1\Leftrightarrow x=0\) (loại)

Vậy ko có giá trị nào của x thỏa mãn yêu cầu bài toán.

c) Yêu cầu bài toán: \(\dfrac{x-1}{2}=1\Leftrightarrow x-1=2\Leftrightarrow x=3\) (nhận)

Vậy \(x=3\) thỏa mãn yêu cầu bài toán.

d) Yêu cầu bài toán: \(\dfrac{x-1}{2}=-3\Leftrightarrow x-1=-6\Leftrightarrow x=-5\) (loại)

Vậy ko có giá trị nào của x thỏa mãn yêu cầu bài toán.

b) \(x^2-y^2+2y-1=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)

câu b sai rồi kìa

B

c. Ta có: \(A=\left(x^3-8\right).\dfrac{1}{x-2}+x\)

\(=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x-2}+x\)

\(=\left(x^2+2x+4\right)+x\)

\(=x^2+3x+4\)

\(=\left(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)

\(=\left(x+1,5\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=-1,5\)

Vậy \(MinA=\dfrac{7}{4}\Leftrightarrow x=-1,5\)

5. the worst

6. luckier

a. Điều kiện xác định: \(\left\{{}\begin{matrix}x^2-4\ne0\\6-3x\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

\(G=\left(\dfrac{x}{x^2-4}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\dfrac{6}{x+2}\)

\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\right]:\dfrac{6}{x+2}\)

\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{6}{x+2}\)

\(=\dfrac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)

\(=\dfrac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-1}{x-2}\)

b. Theo giả thiết, ta có:

\(\left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

So với điều kiện xác định thấy \(x=-1\) thỏa mãn.

Thay \(x=-1\) vào biểu thức \(G\) ta có: 

\(G=\dfrac{-1}{-1-2}=\dfrac{-1}{-3}=\dfrac{1}{3}\)