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Trân
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Midoriya Izuku
12 tháng 11 2023 lúc 19:08

9. A
10. D
5. A
6. B
7. D
8. B
 

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Sinh Viên NEU
13 tháng 11 2023 lúc 0:06

9 A

10 D

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Trân
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Nguyễn Lê Phước Thịnh
14 tháng 11 2023 lúc 17:36

e: \(\dfrac{x^2+3x+9}{x^3+4x^2+4x}\cdot\dfrac{x^2+2x}{x^3-27x}\)

\(=\dfrac{x^2+3x+9}{x\left(x^2+4x+4\right)}\cdot\dfrac{x\left(x+2\right)}{x\left(x^2-27\right)}\)

\(=\dfrac{x^2+3x+9}{\left(x+2\right)^2}\cdot\dfrac{x+2}{x\left(x^2-27\right)}\)

\(=\dfrac{\left(x^2+3x+9\right)}{\left(x+2\right)\cdot x\left(x^2-27\right)}\)

f: \(\dfrac{2x^2+4xy+2y^2}{5x-5y}\cdot\dfrac{15x-15y}{2x^3+2y^3}\)

\(=\dfrac{2\left(x^2+2xy+y^2\right)}{5\left(x-y\right)}\cdot\dfrac{15\left(x-y\right)}{2\left(x^3+y^3\right)}\)

\(=\dfrac{\left(x+y\right)^2}{1}\cdot\dfrac{3}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)

\(=\dfrac{3\left(x+y\right)}{x^2-xy+y^2}\)

g: \(\dfrac{x^3-4x}{x^2-7x+12}\cdot\dfrac{x-4}{x^2-2x}\)

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

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

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kiên trần
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Trân
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Toru
5 tháng 9 2023 lúc 12:12

4. \(x^2-3x+xy-3y=0\)

\(\Leftrightarrow x\left(x-3\right)+y\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-y\end{matrix}\right.\)

5. \(x^2-8x-3x+24=0\)

\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)

6. \(\left(x-2\right)^2-5\left(2-x\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

7. \(3x\left(x-1\right)-x^2+2x-1=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[3x-\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

8. \(x^2\left(x-3\right)+18-6x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\pm\sqrt{6}\end{matrix}\right.\)

10. \(\left(x-5\right)^2-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[\left(x-5\right)-\left(x-2\right)\right]\left[\left(x-5\right)+\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5-x+2\right)\left(x-5+x-2\right)=0\)

\(\Leftrightarrow-3\left(2x-7\right)=0\)

\(\Leftrightarrow2x-7=0\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

12. \(x^2\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\)

14. \(3x^2-7x-10=0\)

\(\Leftrightarrow3x^2+3x-10x-10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

#Urushi

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Nguyễn Lê Phước Thịnh
5 tháng 9 2023 lúc 12:14

4: x^2-3x+xy-3y=0

=>x(x-3)+y(x-3)=0

=>(x-3)(x+y)=0

=>x=3 và x+y=0

=>x=3 và y=-3

6: (x-2)^2-5(2-x)=0

=>(x-2)^2+5(x-2)=0

=>(x-2)(x-2+5)=0

=>(x-2)(x+3)=0

=>x=-3 hoặc x=2

8: x^2(x-3)+18-6x=0

=>x^2(x-3)-6(x-3)=0

=>(x-3)(x^2-6)=0

=>x=3 hoặc \(x=\pm\sqrt{6}\)

10: (x-5)^2-(x-2)^2=0

=>(x-5-x+2)(x-5+x-2)=0

=>-3(2x-7)=0

=>2x-7=0

=>x=7/2

12: x^2(x-3)-4x+12=0

=>x^2(x-3)-4(x-3)=0

=>(x-3)(x^2-4)=0

=>(x-3)(x-2)(x+2)=0

=>\(x\in\left\{3;2;-2\right\}\)

14: 3x^2-7x-10=0

=>3x^2-10x+3x-10=0

=>(3x-10)(x+1)=0

=>x=10/3 hoặc x=-1

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18. Ngô Thị Ái Ngọc
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TnLt
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Phạm Nhi
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kodo sinichi
28 tháng 3 2022 lúc 5:41

D

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Khôi Bùi
28 tháng 3 2022 lúc 18:11

\(lim\dfrac{2\sqrt{7n^2-2n}}{3n+2}=lim\dfrac{2\sqrt{n^2\left(7-\dfrac{2}{n}\right)}}{3n+2}=lim\dfrac{2n\sqrt{7-\dfrac{2}{n}}}{n\left(3+\dfrac{2}{n}\right)}\)

\(=lim\dfrac{2\sqrt{7-\dfrac{2}{n}}}{3+\dfrac{2}{n}}=\dfrac{2\sqrt{7}}{3}\) \(=\dfrac{a\sqrt{7}}{b}\) 

Suy ra : a/b = 2/3 => a - b = -1 

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Trương Thuỳ
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Thư Thư
27 tháng 6 2023 lúc 12:14

\(\dfrac{x-1}{x+2}+\dfrac{6x}{x^2-4}=\dfrac{x+1}{2-x}\left(dkxd:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-1}{x+2}+\dfrac{6x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{x+1}{x-2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+6x+\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2-2x-x+2+6x+x^2+2x+x+2=0\)

\(\Leftrightarrow2x^2+6x+4=0\)

\(\Leftrightarrow2x^2+2x+4x+4=0\)

\(\Leftrightarrow2x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-1\right\}\)

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Lê Tuấn
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Đăng Khoa
30 tháng 12 2023 lúc 11:34

Cậu nên tách ra và đăng từng bài nha

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Sinh Viên NEU
31 tháng 12 2023 lúc 9:17

III

1 Did you attend the Ban Flower Festival in Dien Bien last year?

2 Is Mua Sap a popular folk dance of the Thai people?

IV

1 They live in Wewbley, in north London

2 Her name is Tracy

3 No, there isn't

4 There are 4 people in his family

V

1  - B

2 - C 

3 - d

4 - f

5 - a

6 - e

7 - g

 

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