Bạn nào cho mk 1 ik, mk cho bn ý 3 ik luôn. Mk hứa nếu nói dối bạn có thể Báo cáo sai phạm mk.

\(\frac{2}{x-1}+\frac{5}{x+2}=\frac{13}{x^2+x-2}.\)

\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{5\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=\frac{13}{x^2+x-2}\)

\(\Leftrightarrow\frac{2x+4}{x^2+x-2}+\frac{5x-5}{x^2+x-2}=\frac{13}{x^2+x-2}\)

\(\Leftrightarrow\frac{7x-1}{x^2+x-2}=\frac{13}{x^2+x-2}\)

\(\Leftrightarrow7x-1=13\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\)

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

\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=\left(x-3\right)\left(6x-1\right)\)

\(\Leftrightarrow6x^2+4x-3x-2=6x^2-x-18x+3\)

\(\Leftrightarrow4x-3x+x+18x=3+2\)

\(\Leftrightarrow20x=5\)

\(\Leftrightarrow x=\frac{1}{4}\)

1D 2C 3B 5B 6B

7C 8D 10D 12B 16B

17A 19C 20B 21C 22A

1.D

2.C

3. B 

5. B

6. B

7. C

8. D

9. B

10. D

12. B

16. B

17. A

18. B

19. C

20. B

21. C

22. A

h) Ta có: \(\left\{{}\begin{matrix}\left|x-7\right|=\left|7-x\right|\ge7-x\\\left|x+5\right|\ge x+5\end{matrix}\right.\)

\(\Rightarrow\left|7-x\right|+\left|x+5\right|\ge\left(7-x\right)+\left(x+5\right)\)

\(\Rightarrow\left|x-7\right|+\left|x+5\right|\ge12\)

\(\Rightarrow H\ge12\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}7-x\ge0\\x+5\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le7\\x\ge-5\end{matrix}\right.\)

\(\Leftrightarrow-5\le x\le7\)

Vậy, Min_H = 12 \(\Leftrightarrow-5\le x\le7\)

a) Ta có: \(A=2x^2-8x+10\)

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

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

\(2\left[\left(x-2\right)^2+1\right]\)

Ta lại có: \(\left(x-2\right)^2\ge0\)

\(\Rightarrow2\left[\left(x-2\right)^2+1\right]\ge2\)

\(\Rightarrow A\ge2\)

Dấu bằng xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy Min_A = 2 \(\Leftrightarrow x=2\)

a) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

b) \(x^2+12xy+36y^2=\left(x+6y\right)^2\)

c) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

d) Không phải hằng đẳng thức \(\left(x^2-2x+4=\left(x-1\right)^2+3\right)\)

e) \(25x^2+4y^2-20xy=\left(5x-2y\right)^2\)

a. \(\dfrac{\left(x^2+2x\right)}{\left(x+2\right)^2}=\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)

b. \(\dfrac{x^2-7x+12}{x^2-6x+9}=\dfrac{x^2-3x-4x+12}{\left(x-3\right)^2}\)

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

c. \(\dfrac{x^2-5x+6}{x^2-x-2}=\dfrac{x^2-2x-3x+6}{x^2-2x+x-2}\)

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

d. \(\dfrac{\left(x+y\right)^2-z^2}{2\left(x+y+z\right)}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{2\left(x+y+z\right)}=\dfrac{x+y-z}{2}\)

\(a,\)Mình làm theo kiểu lược đồ

Nhẩm nghiệm của đa thức trên ta đc : 2

Có lược đồ sau :(dòng trên ghi các hệ số)

Ta phân tích đc thành :\(\left(x-2\right)\left(x^2-6\right)\)

\(c,x^2-5x+4\)

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

\(=x\left(x-4\right)-\left(x-4\right)\)

\(=\left(x-1\right)\left(x-4\right)\)

\(d,3x^2+5x+2\)

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

\(=3x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(3x+2\right)\)

\(e,x^3-x+3x^2y+3xy^2+y^3-y\)

\(=\left(x^3+y^3\right)+3xy\left(x+y\right)-\left(x+y\right)\)

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

\(=\left(x+y\right)\left[\left(x^2-xy+y^2\right)+3xy-1\right]\)

\(x^3-2x^2-6x+12\)

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

\(=\left(x-2\right)\left(x^2-6\right)\)

\(x^4-7x^2+12\)

\(=\left[\left(x^2\right)^2-2.3,5x+3,5^2\right]-0,25\)

\(=\left(x^2-3,5\right)^2-0,5^2\)

\(=\left(x^2-3,5-0,5\right)\left(x^2-3,5+0,5\right)\)

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

Câu c tương tự câu b