b) \(\Leftrightarrow3x^3+12x-2x^2-8=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(12x-8\right)=0\\ \Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\\ \Leftrightarrow\left(x^2+4\right)\left(3x-2\right)=0\)

Vì \(x^2+4>0\Rightarrow3x-2=0\Rightarrow x=\dfrac{2}{3}\)

c) \(x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=36\\ \Leftrightarrow x^3+4x-x^3=63\\ \Leftrightarrow4x=63\\ \Leftrightarrow x=\dfrac{63}{4}\)

b) 3x(x\(^3\) +12x-2x\(^2\)-8=0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}X^2+4=0\\3X-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x\in Z\\X=\dfrac{2}{3}\end{matrix}\right.\)
 

a) x\(^2\)+5x=0

x(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
 

c)(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=36

x\(^3\)-27+x(x+2)(2-x)=36

4x-27=36

4x=36+27

4x=63

x=\(\dfrac{63}{4}\)

\(36x^2-49=0\)

\(\Leftrightarrow36x^2=49\)

\(\Leftrightarrow x=\sqrt{\dfrac{49}{36}}=\dfrac{7}{6}\)

Vậy: \(x=\dfrac{7}{6}\)

\((6x)^2-7^2=0\)

(6x-7)(6x+7)=0

Th1 6x-7=0

        X=7/6

Th2 6x+7=0

       X=-7/6

Pt có tập nghiệm S=7/6;-7/6

Ta có: \(36x^2-49=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

`x^2=3`

`=>x=\sqrt{3}\or\x=-\sqrt{3}`

`x^2=36`

`<=>x^2=(+-6)^2`

`<=>x=+-6`

`x^2=25`

`<=>x^2=(+-5)^2`

`<=>x=+-5`

`2x^2+(-20)=55`

`<=>2x^2-20=55`

`<=>2x^2=75`

`<=>x^2=75/2`

`<=>x=+-\sqrt{75/2}`

`2(x-1)^2+5^0=9`

`<=>2(x-1)^2+1=9`

`<=>2(x-1)^2=8`

`<=>(x-1)^2=4`

`<=>x-1=2\or\x-1=-2`

`<=>x=3\or\x=-1`

Bài 1.

a: Ta có: \(2x\left(x-3\right)+x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

b: -7<x<7

Bạn coi lại đề. Dấu bị lỗi.

Bài 3:

b. $B=(x+y)(2x-y)+(xy^4-x^2y^2):(xy^2)$

$=(2x^2-xy+2xy-y^2)+(y^2-x)$

$=2x^2+xy-y^2+y^2-x=2x^2+xy-x$

Bài 4:
a. $25x^3-10x^2+x=x(25x^2-10x+1)=x(5x-1)^2$
b. $x^2-9x+9y-y^2=(x^2-y^2)-(9x-9y)=(x-y)(x+y)-9(x-y)=(x-y)(x+y-9)$

c. $16-x^2-4y^2-4xy=16-(x^2+4y^2+4xy)$

$=4^2-(x+2y)^2=(4-x-2y)(4+x+2y)$

Bài 5:

a. $36-x^3=100$

$x^3=36-100=-64=(-4)^3$

$\Rightarrow x=-4$

b.

$(x+2)(x-2)-(x+1)^2=7$

$\Leftrightarrow (x^2-4)-(x^2+2x+1)=7$

$\Leftrightarrow -2x-5=7$

$\Leftrightarrow -2x=12$

$\Leftrightarrow x=-6$