a, Thắc mắc đề cóa sai khong .

( đáp án vẫn có nhưng là số vô tỉ nên nghe lạ á )

b, Ta có : \(x^3-12x^2+48x-72=0\)

=> \(x^3-3.x^2.4+3.x.4^2-64-8=0\)

=> \(\left(x-4\right)^3-8=0\)

=> \(\sqrt[3]{\left(x-4\right)^3}=\sqrt[3]{8}=2\)

=> \(x=6\)

Vậy ....

a) Ta có: \(x^3+3x^2+3x+2=0\)

\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)

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

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

mà \(x^2+x+1\ne0\forall x\)

nên x+2=0

hay x=-2

Vậy: x=-2

b) Ta có: \(x^3-12x^2+48x-72=0\)

\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)

\(\Leftrightarrow x^2\left(x-6\right)-6x\left(x-6\right)+12\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-6x+12\right)=0\)

mà \(x^2-6x+12\ne0\forall x\)

nên x-6=0

hay x=6

Vậy: x=6

Bài 1:

a) Ta có: \(x^3+3x^2+3x+2=0\)

\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)

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

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

mà \(x^2+x+1>0\forall x\)

nên x+2=0

hay x=-2

Vậy: x=-2

b) Ta có: \(x^3-12x^2+48x-72=0\)

\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)

\(\Leftrightarrow x^2\left(x-6\right)-6x\left(x-6\right)+12\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-6x+12\right)=0\)

mà \(x^2-6x+12>0\forall x\)

nên x-6=0

hay x=6

Vậy: x=6

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

x³ + 12x² + 48x + 64 = 8x³ - 12x² + 6x - 1

\(\Leftrightarrow\) x³ + 12x² + 48x + 64 - 8x³ + 12x² - 6x + 1 = 0

\(\Leftrightarrow\) -7x³ + 24x² + 42x + 65 = 0

Bn cho đề thế này ai mà giải được :vvv

bằng (x + 4)³

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

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

=>(x-4)^3=0

=>x-4=0

=>x=4

    \(x^3+12x^2+48x+64\)

= \(\left(x^3+64\right)+\left(12x^2+48x\right)\)

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

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

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

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

pt<=>\(\sqrt{\left(x+6\right)^3}+\sqrt{x+6}=\left(x^2+4x\right)^3+x^2+4x\)

đặt\(\sqrt{x+6}=a;x^2+4x=b\)