2/2 x 4 + 2/4x 6 + 2/ 6x 8 + 2/ 2014 x 2016
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\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2014.2016}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{2014}-\frac{1}{2016}\)
\(=\frac{1}{2}-\frac{1}{2016}\)
\(=\frac{1008}{2016}-\frac{1}{2016}=\frac{1007}{2016}\)
\(\frac{2}{2\times4}+\frac{2}{4\times6}+...+\frac{2}{2014\times2016}\)
=\(\left(\frac{2}{2}-\frac{2}{4}\right)+\left(\frac{2}{4}-\frac{2}{6}\right)+...+\left(\frac{2}{2014}-\frac{2}{2016}\right)\)
=\(\frac{2}{2}-\frac{2}{4}+\frac{2}{4}-\frac{2}{6}+...+\frac{2}{2014}-\frac{2}{2016}\)
=\(\frac{2}{2}-\frac{2}{2016}=\frac{1007}{1008}\)

Bài 1:
\(=\left(3x-1\right)^2-9y^2\)
=(3x-1-3y)(3x-1+3y)
=(3x−1)2−9y2=(3x−1)2−9y2
=(3x-1-3y)(3x-1+3y)
Tham khảo ạ

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 \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)

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.