\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2014.2016}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{2014}-\frac{1}{2016}\)

\(=\frac{1}{2}-\frac{1}{2016}\)

\(=\frac{1008}{2016}-\frac{1}{2016}=\frac{1007}{2016}\)

\(\frac{2}{2\times4}+\frac{2}{4\times6}+...+\frac{2}{2014\times2016}\)

=\(\left(\frac{2}{2}-\frac{2}{4}\right)+\left(\frac{2}{4}-\frac{2}{6}\right)+...+\left(\frac{2}{2014}-\frac{2}{2016}\right)\)

=\(\frac{2}{2}-\frac{2}{4}+\frac{2}{4}-\frac{2}{6}+...+\frac{2}{2014}-\frac{2}{2016}\)

=\(\frac{2}{2}-\frac{2}{2016}=\frac{1007}{1008}\)

Bài 1:

\(=\left(3x-1\right)^2-9y^2\)

=(3x-1-3y)(3x-1+3y)

=(3x-1-3y)(3x-1+3y)
Tham khảo ạ

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

a: ĐKXĐ: \(x^2-6x+6\ge0\)

=>\(x^2-6x+9-3\ge0\)

=>\(\left(x-3\right)^2-3\ge0\)

=>\(\left(x-3\right)^2\ge3\)

=>\(\left[\begin{array}{l}x-3\ge\sqrt3\\ x-3\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\sqrt3+3\\ x\le-\sqrt3+3\end{array}\right.\)

Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

=>\(x^2-6x+6-4\cdot\sqrt{x^2-6x+6}+3=0\)

=>\(\left(\sqrt{x^2-6x+6}-3\right)\left(\sqrt{x^2-6x+6}-1\right)=0\)

TH1: \(\sqrt{x^2-6x+6}-3=0\)

=>\(\sqrt{x^2-6x+6}=3\)

=>\(x^2-6x+6=9\)

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

=>\(x^2-6x+9-12=0\)

=>\(\left(x-3\right)^2=12\)

=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3+3\left(nhận\right)\\ x=3-2\sqrt3\left(nhận\right)\end{array}\right.\)

TH2: \(\sqrt{x^2-6x+6}-1=0\)

=>\(x^2-6x+6=1\)

=>\(x^2-6x+5=0\)

=>(x-1)(x-5)=0

=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

\(x^2-x+8-4\sqrt{x^2-x+4}=0\)

=>\(x^2-x+4-4\cdot\sqrt{x^2-x+4}+4=0\)

=>\(\left(\sqrt{x^2-x+4}-2\right)^2=0\)

=>\(\sqrt{x^2-x+4}-2=0\)

=>\(\sqrt{x^2-x+4}=2\)

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

=>\(x^2-x=0\)

=>x(x-1)=0

=>x=0 hoặc x=1

c: \(x^2+\sqrt{4x^2-12x+44}=3x+4\)

=>\(x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

=>\(x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

=>\(\left(\sqrt{x^2-3x+11}+5\right)\left(\sqrt{x^2-3x+11}-3\right)=0\)

=>\(\sqrt{x^2-3x+11}-3=0\)

=>\(\sqrt{x^2-3x+11}=3\)

=>\(x^2-3x+11=9\)

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

=>(x-1)(x-2)=0

=>x=1(nhận) hoặc x=2(nhận)

1: \(x^2+3x+2\)

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

=x(x+1)+2(x+1)

=(x+1)(x+2)

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

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

=x(x+1)+3(x+1)

=(x+1)(x+3)

3: \(x^2+5x+4\)

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

=x(x+1)+4(x+1)

=(x+1)(x+4)

4: \(x^2-4x+3\)

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

=x(x-1)-3(x-1)

=(x-1)(x-3)

5: \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

6: \(x^2-5x+4\)

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

=x(x-1)-4(x-1)

=(x-1)(x-4)

7: \(x^2-5x+6\)

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

=x(x-2)-3(x-2)

=(x-2)(x-3)

8: \(x^2+6x+5\)

\(=x^2+x+5x+5\)

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

=(x+1)(x+5)

9: \(x^2-7x+10\)

\(=x^2-2x-5x+10\)

=x(x-2)-5(x-2)

=(x-2)(x-5)

10: \(x^2+8x+12\)

\(=x^2+2x+6x+12\)

=x(x+2)+6(x+2)

=(x+2)(x+6)

11: \(x^2-8x+16=x^2-2\cdot x\cdot4+4^2=\left(x-4\right)^2\)

12: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

=x(x+3)+5(x+3)

=(x+3)(x+5)

13: \(x^2-8x+7\)

\(=x^2-x-7x+7\)

=x(x-1)-7(x-1)

=(x-1)(x-7)

14: \(x^2+9x+8\)

\(=x^2+x+8x+8\)

=x(x+1)+8(x+1)

=(x+1)(x+8)

15: \(x^2-9x+14\)

\(=x^2-2x-7x+14\)

=x(x-2)-7(x-2)

=(x-2)(x-7)

16: \(x^2+9x+18\)

\(=x^2+3x+6x+18\)

=x(x+3)+6(x+3)

=(x+3)(x+6)

17: \(x^2-9x+20\)

\(=x^2-4x-5x+20\)

=x(x-4)-5(x-4)

=(x-4)(x-5)

18: \(2x^2-3x+1\)

\(=2x^2-2x-x+1\)

=2x(x-1)-(x-1)

=(x-1)(2x-1)

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

2. \(x^2+4x+3=\left(x+1\right)\left(x+3\right)\)

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

4. \(x^2-4x+3=\left(x-1\right)\left(x-3\right)\)

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

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

7. \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

8. \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)

9. \(x^2-7x+10=\left(x-2\right)\left(x-5\right)\)

10. \(x^2+8x+12=\left(x+2\right)\left(x+6\right)\)

11. \(x^2-8x+16=\left(x-4\right)^2\)

12. \(x^2+8x+15=\left(x+3\right)\left(x+5\right)\)

13. \(x^2-8x+7=\left(x-1\right)\left(x-7\right)\)

14. \(x^2+9x+8=\left(x+1\right)\left(x+8\right)\)

15. \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)

16. \(x^2+9x+18=\left(x+3\right)\left(x+6\right)\)

17. \(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(18.2x^2-3x+1=2x^2-x-2x+1\)

\(=x\cdot\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(x-1\right)\)

Ko chắc nhá, lúc làm chả biết có tính nhầm chỗ nào ko nữa:) Vả lại bài này chưa khảo lại bài đâu đấy, lười khảo lại lắm, đăng lên luôn.

a) ĐK: \(x\ge-\frac{1}{4}\)

PT \(\Leftrightarrow4x^2+4x+1-2\sqrt{4x+1}+1=0\)

\(\Leftrightarrow4x^2+\left(\sqrt{4x+1}-1\right)^2=0\)

b) ĐK: \(x\ge-\frac{1}{2}\)

PT \(\Leftrightarrow\left(x^2-8x+16\right)+2x+1-6\sqrt{2x+1}+9=0\)

\(\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{2x+1}-3\right)^2=0\)

c) ĐK: \(x\ge-1\)

PT có một nghiệm xấu @@ chưa nghĩ ra, có lẽ phải dùng liên hợp.

d) Số bự quá:( Nhưng thôi vì nghiệm đẹp nên vẫn làm:D

\(PT\Leftrightarrow\left(x^2-2x+1\right)+\left(2017x-2016-2\sqrt{2017x-2016}+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2017x-2016}-1\right)^2=0\)

e)Nghiệm đẹp nhưng dạng phân thức -> ko muốn làm:D

f) Liên hợp đi cho nó khỏe:v

f) Liên hợp đi cho nó khỏe:D

ĐK: \(x\ge\frac{1}{5}\)

PT \(\Leftrightarrow2x^2-6x+4+\left(x+1\right)-\sqrt{5x-1}=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)+\frac{\left(x-2\right)\left(x-1\right)}{x+1+\sqrt{5x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left[2+\frac{1}{x+1+\sqrt{5x-1}}\right]=0\)

Cái ngoặc to nhìn liếc qua một phát cũng thấy nó vô nghiệm.

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.