a) \(x^6-y^6=\text{(x-y)(y+x)(y^2-xy+x^2)(y^2+xy+x^2)}\)                                                                                                            b)\(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

a)\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

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

2. Ta có : \(\hept{\begin{cases}a+b+c=2016\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2016}\end{cases}\Leftrightarrow}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow a+b=0\) hoặc \(b+c=0\) hoặc \(a+c=0\)

• Nếu a + b = 0 => c = 2016 (1)
• Nếu b + c = 0 => a = 2016 (2)
• Nếu a + c = 0 => b = 2016 (3)

Từ (1) , (2) , (3)  ta có điều phải chứng minh.

Gọi hai số chính phương liên tiếp lần lượt là \(n^2,\left(n+1\right)^2\) (\(n\in N^{\text{*}}\))

Ta có : \(n^2+\left(n+1\right)^2+n^2\left(n+1\right)^2=n^2\left(n+1\right)^2+n^2+\left(n^2+2n+1\right)\)

\(=n^2\left(n+1\right)^2+2n\left(n+1\right)+1=\left[n\left(n+1\right)+1\right]^2\)

Dễ thấy n(n+1) chia hết cho 2 vì là tích của hai số tự nhiên liên tiếp => n(n+1) là số chẵn => n(n+1) + 1 là số lẻ

\(\Rightarrow\left[n\left(n+1\right)+1\right]^2\) là một số chính phương lẻ.

Vậy ta có điều phải chứng minh.

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

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

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

\(=\left(x^4-x^2+1\right)^2\ge0\)

\(c^2+2\left(ab-bc-ac\right)\Leftrightarrow-c^2=\left(ab-bc-ac\right)\)

Ta có : \(2a^2-2ac+c^2=a^2-c^2+c^2+\left(a-c\right)^2=a^2+c^2+2\left(ab-bc-ac\right)+\left(a-c\right)^2\)

\(=\left(a^2-2ac+c^2\right)+2b\left(a-c\right)+\left(a-c\right)^2=\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2\)

\(=\left(a-c\right)\left(2a-2c+2b\right)=2\left(a-c\right)\left(a+b-c\right)\)

Tương tự ở mẫu số ta cũng có : \(2b^2-2bc+c^2=2\left(b-c\right)\left(a+b-c\right)\)

\(\Rightarrow\frac{2a^2-2ac+c^2}{2b^2-2bc+c^2}=\frac{2\left(a-c\right)\left(a+b-c\right)}{2\left(b-c\right)\left(a+b-c\right)}=\frac{a-c}{b-c}\)

Bít chết liền

5x(1 - 2x) - 3x(x + 18) = 0

=> 5x - 10x² - 3x² - 54x = 0

=> -13x² - 49x = 0

=> x.(-13x - 49) = 0

=> x = 0

hoặc -13x - 49 = 0 => -13x = 49 => x = -49/13

                                                                      Vậy x = 0, x = -49/13

\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)

\(\Rightarrow x\left(5\left(1-2x\right)-3\left(x+18\right)\right)=0\)

\(\Rightarrow x\left(5-10x-3x+54\right)=0\)

\(\Rightarrow x\left(59-13x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\59-13x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\13x=59\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\frac{59}{13}\end{cases}}}\)

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

\(=-3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}+\frac{2}{9}\right)\)

\(=-3\left(\left(x-\frac{1}{3}\right)^2+\frac{2}{9}\right)\)

\(=\frac{-2}{3}-3\left(x-\frac{1}{3}\right)^2\le\frac{-2}{3}\)

MIN A \(=\frac{-2}{3}\Leftrightarrow x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)

\(2x-3x^2-1=-3\left(x^2-2.x.\frac{1}{3}+\frac{1}{9}\right)+\frac{1}{3}-1=-3\left(x-\frac{1}{3}\right)^2-\frac{2}{3}\le-\frac{2}{3}\)

Vậy Max = \(-\frac{2}{3}\Leftrightarrow x=\frac{1}{3}\)