Đề là gì bn

1. x² - 4x + 4 - x²y + 2xy

= (x - 2)² - xy(x - 2)

= (x - 2 - xy)(x - 2)

2. x² + 4x - 5

= x² - x + 5x - 5

= x(x - 1) + 5(x - 1)

= (x + 5)(x - 1)

3. x² + 7x + 12

= x² + 3x + 4x + 12 

= x(x + 3) + 4(x + 3)

= (x + 4)(x + 3)

1) \(x^2-4x+4-x^2y+2xy=\left(x-2\right)^2-xy\left(x-2\right)=\left(x-2\right)\left(x-2-xy\right)\)

2) \(x^2+4x-5=x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(x+5\right)\)

3) \(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

Bài nào -.-

Viết lại câu:

6 A

7 A

8 D

9 B

10 C

11 D 

12 A

13 B

14 D

15 A

Lỗi sai

1 C => which

2 C => stopped

3 A => bored

4 B => most

5 B => had stopped

6 C => that

7 C => had knowm

8 A => hadn't been

9 D => learned about

10 B => bỏ

11 B => have come

12 A => would have gone

13 C => would have returned

14 C => knew

6.a

7.a

8.d

9.b

10.c

11.d

12.a

13.b

14.d

15.a

Bài chọn lỗi sai

1.c =>which

2.c => stopped

3.a =>bored

4.b => most

5.b =>had stopped

6.c =>that

7.c => had known

8.a => hadn't been late

9.d => learned about

10.b => rỗng

11.b => would have come

12.a => would have gone

13.c => would have returned

14.c => knew

Lời khuyên chân thành là em nên đọc kĩ bài, phần cô giáo phân tích trên lớp để làm, như vậy em vừa hiểu kĩ bài vừa là tự em sẽ rèn được kĩ năng làm bảng

....."màu xanh".... là sai

\(\dfrac{2x-3}{4}.\dfrac{6}{5}=\dfrac{21}{10}\\ \Leftrightarrow\dfrac{2x-3}{4}=\dfrac{7}{4}\\ \Leftrightarrow2x-3=7\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\)

x = 5

x = 5

Bài 6

\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)

Bài 7:

\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)

Điều kiện (x≠5, x≠-5)

\(P=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

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

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

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