Ta có

x 3   –   12 x 2   +   48 x   –   64   =   0     ⇔   x 3   –   3 . x 2 . 4   +   3 . x . 4 2   –   4 3   =   0     ⇔   ( x   –   4 ) 3   =   0

ó x – 4 = 0 ó x = 4

Vậy x = 4

Đáp án cần chọn là: B

-1000

\(x^3-12x^2+48x-64=\left(x-4\right)^3=\left(-6-4\right)^3=-10^3=-1000\)

x3 + 12x2 + 48x + 64 = x3 + 3.x2.4 + 3.x.42 + 43 = (x + 4)3

Tại x = 6, giá trị biểu thức bằng (6 + 4)3 = 103 = 1000.

Ta có x 3   +   12 x 2   +   48 x   +   64   =   x 3   +   3 x 2 . 4   +   3 . x . 4 2   +   4 3   =   ( x   +   4 ) 3

Đáp án cần chọn là: A

1:

a: A=(x+4)^3=10^3=1000

b: B=(x-2)^3=20^3=8000

1

a) \(A=x^3+3.x^2.4+3x.4^2+4^3=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

b) \(B=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

2

\(VT=\left(x-y\right)^2+4xy=x^2-2xy+y^2+4xy=x^2+2xy+y^2=\left(x+y\right)^2=VP\)

\(a,=\left(x+3\right)^3=\left(7+3\right)^3=10^3=1000\\ b,=\left(4-x\right)^3=\left(4-24\right)^3=\left(-20\right)^3=-8000\\ c,=\left(x-1\right)^3=\left(11-1\right)^3=10^3=1000\)

a: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

=>\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

=>9x+7=16

=>9x=9

=>x=1

b: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=17\)

=>\(x^3-25x-x^3-8=17\)

=>-25x=8+17=25

=>x=-1

c: \(x^3-12x^2+48x-64=0\)

=>\(\left(x-4\right)^3=0\)

=>x-4=0

=>x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

a: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

=>\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

=>9x+7=16

=>9x=9

=>x=1

b: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=17\)

=>\(x^3-25x-x^3-8=17\)

=>-25x=8+17=25

=>x=-1

c: \(x^3-12x^2+48x-64=0\)

=>\(x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3=0\)

=>\(\left(x-4\right)^3=0\)

=>x-4=0

=>x=4