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Bài 1: Cho parabol (P): y=\(\dfrac{1}{2}x^2\) và đường thẳng (d): y=mx\(-\)2m\(+\)2. Tìm m để đường thẳng (d) cắt parabol (P) tại 2 điểm phân biệt có hoành độ \(x_1,x_2\) sao cho \(x_2=8x_1\).

Bài 2: Cho parabol (P): y= -x²  và đường thẳng (d): y= -mx+m-1. Tìm m để đường thẳng (d) cắt parabol (P) tại 2 điểm phân biệt có hoành độ \(x_1,x_2\) sao cho \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{3}{2}\).

Bài 3: Cho đường thẳng (d): \(y=\left(m^2+1\right)x+2\). Đường thẳng (d) cắt Ox tại A, cắt Oy tại B. Tìm m sao cho khoảng cách từ gốc tọa độ tới đường thẳng (d) lớn nhất.

Giúp mình với, 4h30 mình cần rồi.

1. He made a lot of effort at work. He was not promoted.(in spite of)

=>

2. Athough Tom was a poor student, he studied very well.

=> In spite of

3. In spite of the high prices, my daughter insison goingto the movies.

=> Even though

4. In spite of his good salary, Jack gave up his job.

=> Although

5.Although she read vacancy ads in the newpapers every day, she has not been able to find a job.

=> In spite of

6.After lots effort, the boy was finally successful in getting  the position of striker in the school's football team.

=> After lots of effort. the boy managed

7. It's not a good idea to start your essay without analysing the question first.

=> You should avoid

8. I don't think I will help you with your homework.

=> I don't agree

9. Students will play more active roles in the classroom.

=> More active roles

Không cần làm câu 64 đâu mn.

3) \(\dfrac{A}{B}>1\)

\(\Leftrightarrow\dfrac{x+2\sqrt{x}}{x-1}:\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}>1\)

\(\Leftrightarrow\dfrac{x+2\sqrt{x}}{x-1}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+2}>1\)

\(\Leftrightarrow\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+2}>1\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}.\dfrac{\sqrt{x}}{1}>1\)

\(\Leftrightarrow\dfrac{x}{\sqrt{x}+1}-1>0\)

\(\Leftrightarrow\dfrac{x}{\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow x-\sqrt{x}+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)

\(\Leftrightarrow\sqrt{x}-1>0\)

\(\Leftrightarrow x>1\)

25) \(\dfrac{3x+2}{2}+\dfrac{5-2x}{3}=\dfrac{11}{6}\)

\(\Leftrightarrow\dfrac{9x+6}{6}+\dfrac{10-4x}{6}=\dfrac{11}{6}\)

\(\Leftrightarrow9x+6+10-4x=11\)

\(\Leftrightarrow5x+16=11\)

\(\Leftrightarrow5x=-5\)

\(\Leftrightarrow x=-1\)

26) \(\dfrac{x+2}{4}+\dfrac{2x+3}{3}=\dfrac{x-12}{6}\)

\(\Leftrightarrow\dfrac{3x+6}{12}+\dfrac{8x-12}{12}=\dfrac{2x-24}{12}\)

\(\Leftrightarrow3x+6+8x-12=2x-24\)

\(\Leftrightarrow9x=-18\)

\(\Leftrightarrow x=-2\)

27) \(\dfrac{2x-1}{3}-\dfrac{x-1}{2}+\dfrac{x+1}{6}=1\)

\(\Leftrightarrow\dfrac{4x-2-3x+3+x+1}{6}=1\)

\(\Leftrightarrow4x-2-3x+3+x+1=1\)

\(\Leftrightarrow2x=-1\)

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

Bài 1 tính hoành đọo giao điểm

a) y= a² và y= 2x-1
b) y= a² và y= -x+1