Ta có: \(x^3+12x^2+48x+64=8x^3-12x^2+6x-1\)

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

\(\Leftrightarrow\left(x+2\right)^3-\left(2x-1\right)^3=0\)

\(\Leftrightarrow\left[\left(x+2\right)-\left(2x-1\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(2x-1\right)+\left(2x-1\right)^2\right]=0\)

\(\Leftrightarrow\left(x+2-2x+1\right)\left(x^2+4x+4+2x^2+3x-2+4x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(7x^2+3x-6\right)=0\)

\(\Leftrightarrow7\left(3-x\right)\cdot\left(x^2+\frac{3}{7}x-\frac{6}{7}\right)=0\)

mà 7>0

nên \(\left[{}\begin{matrix}3-x=0\\x^2+\frac{3}{7}x-\frac{6}{7}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+2\cdot x\cdot\frac{3}{14}+\frac{9}{196}-\frac{177}{196}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x+\frac{3}{14}\right)^2=\frac{177}{196}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+\frac{3}{14}=\frac{\sqrt{177}}{14}\\x+\frac{3}{14}=-\frac{\sqrt{177}}{14}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3+\sqrt{177}}{14}\\x=\frac{-3-\sqrt{177}}{14}\end{matrix}\right.\)

Vậy: \(S=\left\{3;\frac{-3+\sqrt{177}}{14};\frac{-3-\sqrt{177}}{14}\right\}\)

bài của bạn có vấn đề gì không vậy?

64-48x+ 12x²- x³

= 4³- 3. 4³.x+ 3x².4- x³

= (4-x)³

Ta có :

\(64-48x+12x^2-x^3\)

\(=4^3-3.4^2x+3.4x^2-x^3\)

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

Bài 4:

a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)

\(=x^2.\left(x+4\right)+8x.\left(x+4\right)+16.\left(x+4\right)\)

\(=\left(x+4\right).\left(x^2+8x+16\right)=\left(x+4\right).\left(x^2+4x+4x+16\right)\)

\(=\left(x+4\right).\left(x+4\right)^2=\left(x+4\right)^3\)(1)

Thay \(x=6\) vào (1) ta được:

\(\left(6+4\right)^3=10^3=1000\)

Vậy...........

b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)

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

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

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

Thay \(x=22\) vào (2) ta được:

\(\left(22-2\right)^3=20^3=8000\)

Vậy.............

Chúc bạn học tốt!!!

Bài 2:

a, \(\left(x+9\right)^3=27=3^3\)

\(\Rightarrow x+9=3\Rightarrow x=-6\)

Vậy.........

b, \(8-12x-x^3+6x^2=-64\)

\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)

\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)

\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)

\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)

Vậy............

Chúc bạn học tốt!!!

a) Ta có: \(x^3+12x^2+48x+64\)

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

\(=\left(x+4\right)^3\)

b) Ta có: \(x^3-12x^2+48x-64\)

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

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

c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

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

d)Sửa đề: \(x^3-3x^2+3x-1\)

Ta có: \(x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)

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

e) Ta có: \(8-12x+6x^2-x^3\)

\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)

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

f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)

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

\(=\left(\frac{1}{3}-3y\right)^3\)

thanks bạn

Thay x = 6 vào biểu thức ta có

\(6^3+12.6^2+48.6+64=1000\)

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.

\(m,x^3+48x=12x^2+64\)

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

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

\(x=4\)

\(n,x^3-3x^2+3x=1\)

\(x^3-3x^2+3x-1=0\)

\(\left(x-1\right)^3=0\)

\(x=1\)

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

\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16\right)-12x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-4\right)^2=0\)

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

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)