a) (x-2x)²=(-x)²=x²

b) (2x²+3)²=4x⁴+12x²+9

c) (x-2)(x²+2x+4)=x³-2³=x³-8

d) (2x-1)³=(2x)³-3.(2x)².1+3.2x.1²-1³=8x³-12x²+6x-1

Chúc bạn học tốt !

cảm ơn

Cách 1: x³-7x-6=x³+x²-x²-x-6x-6=x²(x+1)-x(x+1)-6(x-1)=(x-1)(x²-x-6)

Cách 2: x³-7x-6=x³-x-6x-6=x(x²-1)-6(x+1)=x(x-1)(x+1)-6(x+1)=(x+1)[x(x-1)-6]=(x+1)(x²-x-6)

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

Chúc bạn học tốt !

\(C=x^2-2x+y^2+4y+8\)

\(C=\left(x^2-2x\right)+\left(y^2+4y\right)+8\)

\(C=\left(x^2-2\cdot x\cdot1+1^2\right)+\left(y^2+2\cdot y\cdot2+2^2\right)+\left(8-1-2^2\right)\)

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

mà (x-1)² và (y+2)² luôn lớn hơn hoặc bằng 0

\(\Rightarrow C\ge3\forall x\)

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

Vậy, Cmin = 3 <=> x = 1; y = -2

thanks b

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 :

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

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

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