Bài 1

\(\hept{\begin{cases}a.x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\\b.x^3+8y^3=\left(x+2y\right)\left(x^2+2xy+4y^2\right)\end{cases}}\)

bài 2

\(\hept{\begin{cases}a.\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\\\left(2y+z\right)\left(4y^2-2yz+z^2=8y^3-z^3\right)\end{cases}}\)

bài 3.

\(\hept{\begin{cases}a.\left(x-3\right)\left(x^2+3x+9\right)-\left(30+x^3\right)=\left(x^2-27\right)-\left(30+x^3\right)=-57\\b.\left(2x-y\right)\left(4x^2+2xy+y^2\right)+y^3=8x^3-y^3+y^3=8x^3\end{cases}}\)

bài 4.

\(a.\left(x+2\right)\left(x^2-2x+4\right)=x^3+2x\Leftrightarrow x^3+8=x^3+2x\)

\(\Leftrightarrow2x=8\Leftrightarrow x=4\)

\(b.\left(3x-2\right)\left(9x^2+6x+4\right)-27x^3=3x-5\)

\(\Leftrightarrow27x^3-8-27x^3=3x-5\Leftrightarrow3x=-3\Leftrightarrow x=-1\)

Bài 1:

a) x³-8=x³-2³=(x-2)(x²+2x+2²)

b) x³+8y³=x³+(2y)³=(x+2y)[x²-2xy+(2y)² ]

Bài 2:

a) (x-3)(x²+3x+9)=x³-3³=x³+(-3)³

b) (2y+z)(4y²-2yz+z²)=(2y)³+x³

Bài 3:

a) (x-3)(x²+3x+9)-(30-x³)

=x³-3³-30-x³

=-27-30=-57

b) (2x-y)(4x²+2xy+y²)+y³

=(2x)³-y³+y³

=8x³

Bài 4:

a) (x+2)(x²-2x+4)=x³+2x

=> x³+2³=x³+2x

=> x³+8=x³+2x

=> x³+8-x³=2x

=>8=2x

=>x=8:2=>x=4

b) (3x-2)(9x²+6x+4)-27x³=3x-5

=>(3x)³-2³-(3x)³=3x-5

=>-8=3x-5

=>-8+5=3x

=>3x=-3=>x=-1

a) x³-0,25x=0

<=> x(x²-0,25)=0

<=> x(x-0,5)(x+0,5)=0

<=>\(\hept{\begin{cases}x=0\\x-0,5=0\\x+0,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\pm0,5\end{cases}}}\)

b) x²-10x=-25

<=> x²-10x+25=0

<=> (x-5)²=0

<=> x-5=0

<=> x=5

c) x²-4x=0

<=> x(x-4)=0

<=>\(\hept{\begin{cases}x=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=4\end{cases}}}\)

d₁) x²-8x+16=0

<=> (x-4)²=0

<=> x-4=0

<=> x=4

d₂) 2x²-4x=0

<=> 2x(x-2)=0

<=>\(\hept{\begin{cases}2x=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\end{cases}}}\)

e) x²-16=0

<=> (x-4)(x+4)=0

<=>\(\hept{\begin{cases}x-4=0\\x+4=0\end{cases}\Leftrightarrow x=\pm4}\)

a) x³-0.25x=0

(=) x(x²-0.25)=0

• x=0 
• (x-0.5)(x+0.5)=0 (=)x=0.5 hoac -0.5

Vay x=0 hoac x=+-0.5

 n²(n + 1)+2n(n + 1) (n ∈ Z)

=n(n+1)(n+2)

Mà n(n+1)(n+2)\(⋮\)6 (tích của 3 số nguyên liên tiếp)

Do đó:  n²(n + 1)+2n(n + 1)\(⋮\)6

n2 ( n + 1 ) + 2n ( n + ) ( n thuộc Z ) 

= N ( n + 1 ) ( n+ 2 ) 

mà n ( n + 1 ) ( n + 2 ) :6 ( tích của 3 số nguyên liên tiếp ) 

do đó : n2 ( n + 1 ) + 2n ( n + 1 ) ; 6

a) x+5x²=0

<=> x(1+5x)=0

<=>\(\hept{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)

b) x+1=(x+1)²

<=> x+1-(x+1)²=0

<=> (x+1)(1-x-1)=0

<=> -x(x+1)=0

<=> \(\hept{\begin{cases}-x=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)

c) x³+x=0

<=> x(x²+1)=0

<=> x=0 (vì x²+1>0)

Ta có : (x - 2)(x² + 4) = x² - 2x 

<=> (x - 2)(x² + 4) = x(x - 2) 

<=> (x - 2)(x² - x + 4) = 0

<=> x - 2 = 0 (vì \(x^2-x+4=\left(x-\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\forall x\))

<=> x = 2

Vậy x = 2 

Ta có : (x + 1)(2x - 3) \(\le\)0

 Xét 2 trường hợp : 

TH1 : \(\hept{\begin{cases}x+1\le0\\2x-3\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\le-1\\x\ge1,5\end{cases}}\left(\text{loại}\right)\)

TH2 : \(\hept{\begin{cases}x+1\ge0\\2x-3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\le1,5\end{cases}}\Leftrightarrow-1\le x\le1,5\)

Vậy \(-1\le x\le1,5\)

 y-x^2y-2xy^2-y^3
=y(1-x^2+2xy-y^2)
=y[1^2-(x-y)^2]
=y(1-x+y)(1+x-y)

x² - 2xy + y² - xz + yz

= ( x - y )² - z( x - y )

= ( x - y )( x- y - z )