4x² - 12x = 9

4x² - 12x + 9 = 0

( 2x - 3 )² = 0

2x - 3 = 0

2x = 3

x = 1,5

\(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

\(\Leftrightarrow\left(2x\right)^2-2.2x.3+3^2=0\)

\(\Leftrightarrow\left(2x-3\right)^2=0\)

\(\Leftrightarrow x=1.5\)

vậy...

\(A=4\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^{128}-1\right)< B\)

\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1=B\)

\(\Rightarrow A< B\)

a) \(x^2-6x+8\)

\(=x^2-2\cdot x\cdot3+3^2-1\)

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

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

\(=\left(x-4\right)\left(x-2\right)\)

Còn lại tương tự

a) \(x^2-6x+8=x^2-2x-4x+8\)                     

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

=x(x-2)-4(x-2) = (x-2)(x-4)

       \(a\left(a+1\right)\left(a+2\right)\left(a+3\right)+1\)

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

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

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

\(A=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)

\(=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x=-10\)

Vậy giá trị biểu thức A không phụ thuộc vào biến x

\(A=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)

\(A=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x\)

\(A=\left(5x^2+x^2-6x^2\right)-\left(3x-3x\right)-\left(x^3-x^3\right)-10\)

\(A=0-0-0-10\)

\(A=-10\)

Vậy với mọi x thì A = -10

=> Giá trị của A không phụ thuộc vào biến x

       \(-3x^4y+6x^3y-3x^2y\)

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

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