1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

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

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

Ta có:

$(x+4)^2-x^2(x+12)=16$

$x^2+8x+16-x^3-12x^2=16$

$-x^3-11x^2+8x=0$

$-x(x^2+11x-8)=0$

Suy ra:

$x=0$ hoặc $x^2+11x-8=0$

Giải phương trình bậc hai:

$x=\dfrac{-11\pm\sqrt{121+32}}{2}$

$=\dfrac{-11\pm\sqrt{153}}{2}$

$=\dfrac{-11\pm3\sqrt{17}}{2}$

Vậy: $x=0,\quad x=\dfrac{-11+3\sqrt{17}}2,\quad x=\dfrac{-11-3\sqrt{17}}2$

a) ${\left(2x-1\right)}^2-25=0$

${\left(2x-1\right)}^2=0+25=25$

${\left(2x-1\right)}^2=5^2={\left(-5\right)}^2$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}$

câu b thiếu bn ơi

a: Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

`Answer:`

a, `4x^2-24x+36=(x-3)^3`

`<=>4(x^2-6x+9)-(x-3)^3=0`

`<=>4(x-3)^2-(x-3)^3=0`

`<=>(x-3)^2.(4-x+3)=0`

`<=>(x-3)^2.(7-x)=0`

`<=>x-3=0` hoặc `7-x=0`

`<=>x=3` hoặc `x=7`

b, `(8x^3-7x^2):x^2=3x+\sqrt{\frac{9}{25}}`

`<=>8x^3:x^2-7x^2:x^2=3x+\sqrt{\frac{9}{25}}`

`<=>8x-7=3x+\sqrt{\frac{9}{25}}`

`<=>8x-7=3x+3/5`

`<=>8x=3x+\frac{38}{5}`

`<=>8x-3x=3x+\frac{38}{5}-3x`

`<=>5x=\frac{38}{5}`

`<=>x=\frac{38}{25}`

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

$=x^2(x-5)-(x-5)=0$

$=(x-5)(x^2-1)=0$

$=(x-5)(x-1)(x+1)=0$

$\Rightarrow x=5,\ x=1,\ x=-1$

$3x^4-9x^3+9x^2-27x=0$

$3x(x^3-3x^2+3x-9)=0$

$3x[x^2(x-3)+3(x-3)]=0$

$3x(x-3)(x^2+3)=0$

Vì $x^2+3>0$ nên:

$\Rightarrow x=0$ hoặc $x=3$