a)

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

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

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

\(\Leftrightarrow x+\frac{1}{2}=2x\)

\(\Leftrightarrow x=\frac{1}{2}\)

2)

\(3x^2+6x+100\)

\(=3\left(x^2+2x+\frac{100}{3}\right)\)

\(=3\left(x^2+2\cdot x\cdot1+1^2+\frac{100}{3}\right)\)

\(=3\left[\left(x+1\right)^2+\frac{100}{3}\right]\)

\(=3\left(x+1\right)^2+100\ge100\forall x\left(đpcm\right)\)

a² - a = a ( a - 1 )

mà a và a-1 là 2 số liên tiếp

=> 1 trong 2 số là số chẵn

=> a ( a - 1 ) chia hết cho 2 hay a² - a chia hết cho 2

Ta có : \(a^2-a=a\left(a-1\right)\)

Vì \(a\left(a-1\right)\)là tích 2 số nguyên liên tiếp nên

\(a\left(a-1\right)⋮2\)

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

Vì \(a\left(a-1\right)\left(a+1\right)\)là tích 3 số nguyên liên tiếp nên :

\(a\left(a-1\right)\left(a+1\right)⋮3\)

+ \(a^5-a=a\left(a^4-1\right)\)

\(=a\left(a^2-1\right)\left(a^2+1\right)\)

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

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

\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)\(+5\left(a-1\right)a\left(a+1\right)\)

Vì \(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)là tích 5 số nguyên liên tiếp

\(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)⋮5\)

\(\Rightarrow a^5-a⋮5\)

x³-7x+6=x³-x-6x+6=x(x-1)(x+1)-6(x-1)=(x-1)(x²+x-6)=(x-1)(x-2)(x+3)

\(x^3-7x+6=x^3+3x^2-3x^2-9x+2x+6\)

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

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

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

\(=\left(x+3\right).\left[x.\left(x-1\right)-2.\left(x-1\right)\right]\)

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

Đặt \(a=\frac{1}{315}\),    \(b=\frac{1}{651}\)ta có :

\(A=\left(2+a\right)\cdot b-3a\left(3+1-b\right)-4ab+12a\)

\(\Rightarrow A=2b+ab-12a+3ab-4ab+12a\)

\(\Rightarrow A=2b=\frac{2}{651}\)

\(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)

\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)

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

\(=\left(x+8\right)\left[x.\left(x-2\right)+\left(x-2\right)\right]\)

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

\(x^3-9x^2+6x+16\)

\(=\left(x^3+x^2\right)-\left(10x^2+10x\right)+\left(16x+16\right)\)

\(=x^2.\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-10x+16\right)\)

Tham khảo nhé~

Ta có \(2018-5x^2-y^2-4xy+x\)

\(=2018+\frac{1}{4}-\frac{1}{4}-4x^2-x^2-y^2-2.2xy+x\)

\(=2018+\frac{1}{4}-\left(2x\right)^2-2.2xy-y^2-x^2+2.\frac{1}{2}x-\frac{1}{4}\)

\(=\frac{8073}{4}-\left[\left(2x\right)^2+2.2xy+y^2\right]-\left[x^2-2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right]\)

\(=\frac{8073}{4}-\left(2x+y\right)^2-\left(x-\frac{1}{2}\right)^2\ge\frac{8073}{4}\)( Vì \(\left(2x+y\right)^2\ge0;\left(x-\frac{1}{2}\right)^2\ge0\))

Dấu ''='' xảy ra khi \(\hept{\begin{cases}2x+y=0\\x-\frac{1}{2}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-2x\\x=\frac{1}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}}\)

Vậy GTNN của \(2018-5x^2-y^2-4xy+x\)là \(\frac{8073}{4}\)khi \(x=\frac{1}{2}\)và\(y=-1\)

Ta có : \(a^7-a=a\left(a^6-a\right)\) \(=a\left(a^3-1\right)\left(a^3+1\right)\)

TH1 : \(a⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)\)\(⋮\)\(7\)

TH2 : \(a=7k+1\)\(\left(k\inℤ\right)\)

\(\Rightarrow a^3-1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)\)\(⋮\)\(7\)

TH3 : \(a=7k+2\)\(\left(k\inℤ\right)\)

\(\Rightarrow a^3-1⋮7\)

\(\Rightarrow a^7-a=a\left(a^3+1\right)\left(a^3-1\right)\)\(⋮\)\(7\)

TH4 : \(a=7k+3\)\(\left(k\inℤ\right)\)

\(\Rightarrow a^3+1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)

TH5 : \(a=7k+4\)\(\left(k\inℤ\right)\)

\(\Rightarrow a^3-1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)

TH6 : \(a=7k+5\)\(\left(k\inℤ\right)\)

\(\Rightarrow a^3+1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)

TH7 : \(a=7k+6\)\(\left(k\inℤ\right)\)

\(\Rightarrow a^3+1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)

Vậy trong tất cả các trường hợp ta đều có : \(a^7-a⋮7\)

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=0\)

=> x + y + z = 0

Lai co: x³ + y³ +z³ - 3xyz = (x+y+z).(x²+y²+z² - xy - yz - zx)

             x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

=> 1/xy + 1/yz + 1/xz = 0

=> x + y + z = 0

Lai co: x³ + y³ +z³ - 3xyz = (x+y+z).(x²+y²+z² - xy - yz - zx)

             x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz