1)

x^3 -16x=0`

`<=>x(x^2 -16)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b)

`x^4 -2x^3=0`

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

\(< =>\left[{}\begin{matrix}x^3=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

3)

`(2x-11)(x^2 -1)=0`

\(< =>\left[{}\begin{matrix}2x-11=0\\x^2-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=11\\x^2=1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=1\\x=-1\end{matrix}\right.\)

4)

`x^3 -36x=0`

`<=>x(x^2 -36)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)

5)

`2x+19=0`

`<=>2x=-19`

`<=>x=-19/2`

bài về nghiệm của đa thức

Bài 2: 

a: \(x^2\left(x^2-16\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

a(b+3)-b(3+b)

=(3+b)(a-b)

Thay số, có: (3+1997).(2003-1997)

= 2000.6 =12000

xy(x+y)-2x-2y

xy(x+y)- 2(x+y)

(x+y).(xy-2)

Thay số, co: 7. (8-2)

7.4=28

`4-x=2(x-4)^2`

`<=>4-x=2(x^2-8x+16)`

`<=> 4-x=2x^2 - 16x+32`

`<=> 4-x-2x^2+16x-32=0`

`<=> -2x^2 +15x-28=0`

`<=> -(2x^2-15x+28)=0`

`<=>-(2x^2-7x-8x+28)=0`

`<=> - [x(2x-7) - 4(2x-7)]=0`

`<=> -(2x-7)(x-4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}-2x+7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-7\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

__

`(x^2 +1) (x-2)+2x=4`

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

`<=> x^3 -2x^2 +3x-6=0`

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)

__

`x^4 -16x^2=0`

`<=> x^2 (x^2 -16)=0`

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

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

\(\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)

\(\Leftrightarrow4-x=2x^2-16x+32\)

\(\Leftrightarrow2x^2-15x+28=0\)

\(\Leftrightarrow2x^2-7x-8x+28=0\)

\(\Leftrightarrow x\left(2x-7\right)-4\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7\\x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

___________

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-3\left(\text{vô lý}\right)\\x=2\left(tm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=2\)

________________

\(x^4-16x^2=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 5(8x-5) = 15

<=> 40x = 40 <=> x = 1

Cái này mới chuẩn

b) (2x+1)(1-2x)+(1-2x)²=18 <=> 1 - 4x² + 4x² - 4x + 1 = 18

<=> -4x = 16 <=> x = -4

\(a,\Leftrightarrow x-28=-45\\ \Leftrightarrow x=-27\\ b,\Leftrightarrow3+x=0\\ \Leftrightarrow x=-3\\ c,\Leftrightarrow\left[{}\begin{matrix}7-x=0\\-x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow16\left(x^2-4\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

a)-28+x=-34+(-11)             b)-12(3+x)=0

<=>-28+x=-45                   <=>-36-12x=0

<=>x=-17                            <=>-12x=36

Vậy x=-17                             <=>x=-3

                                         Vậy x=-3

c)(7-x)(-x+2)=0

<=>7-x=0 hoặc -x+2=0

Th1:7-x=0                    Th2:-x+2=0

 <=>x=7                             <=>x=2

           Vậy xϵ{7;2}

d)16x2-64=0

<=>32x=64

<=>x=2

Vậy x=2

bạn tách ra từng ít câu 1 thôi ạ

a: \(=5x\left(xy^2+3x+6y^2\right)\)

b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)

c: \(=\left(x-3\right)\left(x-4\right)\)

d: \(=x\left(x^2-2xy+y^2-9\right)\)

=x(x-y-3)(x-y+3)

e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

f: \(=\left(x-4\right)\left(x+3\right)\)

c: C=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2(x^2-4)

=250x^3+120x-2x^2+8

=250x^3-2x^2+120x+8

d: D=(4x)^3-3^3-(4x)^3-3^3

=64x^3-27-64x^3-27

=-54

c) \(C=\left(5x+2\right)^3+\left(5x-2\right)^3-2\left(x-2\right)\left(x+2\right)\)

\(=\left[\left(5x\right)^3+3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2+2^3\right]+\left[\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2-2^3\right]-2\left(x^2-4\right)\)

\(=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2x^2+8\)

\(=\left(125x^3+125x^3\right)+\left(150x^2-150x^2-2x^2\right)+\left(60x+60x\right)+\left(8-8+8\right)\)

\(=250x^3-2x^2+120x+8\)

d) \(D=\left(4x-3\right)\left(16x^2+12x+9\right)-\left(4x+3\right)\left(16x^2-12x+9\right)\)

\(=\left(4x\right)^3-3^3-\left[\left(4x\right)^3+3^3\right]\)

\(=64x^3-27-\left(64x^3+27\right)\)

\(=64x^3-27-64x^3-27\)

\(=-27-27\)

\(=-54\)