Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

Bài 1: 

a: =42x53+47x42

=42x100

=4200

b: =1152-374-1152-65+374=-65

c: =(-1)+(-1)+...+(-1)+211

=211-105

=106

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

\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=0\)

\(\Leftrightarrow12x+8=0\)

\(\Leftrightarrow12x=-8\)

\(\Leftrightarrow x=-\dfrac{8}{12}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

b) \(\left(2x+3\right)^3-8x\left(x+1\right)\left(x-1\right)=9x\left(4x-3\right)\)

\(\Leftrightarrow8x^3+36x^2+54x+27-8x\left(x^2-1\right)=36x^2-27x\)

\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3+8x=36x^2-27x\)

\(\Leftrightarrow8x^3-8x^3+36x^2-36x^2+54x+27x+8x+27=0\)

\(\Leftrightarrow89x+27=0\)

\(\Leftrightarrow x=-\dfrac{27}{89}\)

c) \(\left(2-x\right)^3+\left(2+x\right)^3-12x\left(x+1\right)=0\)

\(\Leftrightarrow8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x^2-12x=0\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2+6x^2-12x^2\right)-\left(12x-12x\right)+12x+\left(8+8\right)=0\)

\(\Leftrightarrow12x+16=0\)

\(\Leftrightarrow x=-\dfrac{16}{12}\)

\(\Leftrightarrow x=-\dfrac{4}{3}\)

`#040911`

`a)`

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

`<=> x^3 + 6x^2 + 12x + 8 - x^3 - 6x^2 = 0`

`<=> (x^3 - x^3) + (6x^2 - 6x^2) + 12x = 0`

`<=> 12x = 0`

`<=> x = 0`

Vậy, `x = 0.`

`b)`

`(2x + 3)^3 - 8x(x - 1)(x + 1) = 9x(4x - 3)`

`<=> 8x^3 + 36x^2 + 54x + 27 - 8x(x^2 - 1) = 36x^2 - 27x`

`<=> 8x^3 + 36x^2 + 54x + 27 - 8x^3 + 8x - 36x^2 + 27x = 0`

`<=> (8x^3 - 8x^3) + (36x^2 - 36x^2) + (54x + 8x + 27x) + 27 = 0`

`<=> 89x + 27 = 0`

`<=> 89x = -27`

`<=> x = -27/89`

Vậy, `x = -27/89`

`c)`

`(2 - x)^3 + (2 + x)^3 - 12x(x + 1) = 0`

`<=> 8 - 12x + 6x^2 - x^3 + 8 + 12x + 6x^2 + x^3 - 12x^2 - 12x = 0`

`<=> (-x^3 + x^3) + (12x - 12x - 12x) + (6x^2 + 6x^2 - 12x^2) + (8 + 8)=0`

`<=> -12x + 16 = 0`

`<=> -12x = -16`

`<=> 12x = 16`

`<=> x=4/3`

Vậy, `x = 4/3.`

a) |x-2|+|2x+y|≤0

|x-2| ≥ 0

|2x+y|≥ 0

mà |x-2|+|2x+y|≤0⇒|x-2|+|2x+y|=0

⇒|x-2|=0⇒x=2

|2x+y|=0⇒|2.2+y|=0⇒|4+y|=0⇒y=-4

vậy (x,y)={(2;-4)}

a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)

b: -2<x<5

a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

=>\(13\sqrt{2x}=28\)

=>căn 2x=28/13

=>2x=784/169

=>x=392/169

b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>2*căn x-5=4

=>căn x-5=2

=>x-5=4

=>x=9

c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=-1 hoặc x=2

\(b,\left(5x-1\right)^2:2=8\\ \Leftrightarrow\left(5x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\end{matrix}\right.\\ c,\left(1-3x\right)^3=-64\\ \Leftrightarrow1-3x=-4\\ \Leftrightarrow3x=5\\ \Leftrightarrow x=\dfrac{5}{3}\)