ta co: x^3 + 5x -6=x^3-x+6x-6=x.(x^2 - 1) + 6.(x-1) = x.(x-1).(x+1) + 6.(x-1) =(x-1).(x^2+x+6)

Giải thích: `x^2-5x+1`

`=x^2-2. 5/2x+25/4-21/4`

`=(x-5/2)^2-21/4`

`=(x-5/2-\sqrt{21}/2)(x-5/2+\sqrt{21}/2)`

`=(x-[5+\sqrt{21}]/2)(x-[5-\sqrt{21}]/2)`

b. 5x + 5y - 3x - 3y

= (5x - 3x) + (5y - 3y)

= 2x + 2y

= 2(x + y)

Lời giải:

$(x^3-2xy+3x^2)(x-2y)=x(x^2-2xy+3x)(x-2y)$

a, x²-5xy+2x-10y = (x² + 2x)-(5xy+10y)

                            = x(x+2)-5y(x+2)

                            = (x+2)(x-5y)

b, x²-5x+4 = x²- x - 4x +4

                  = (x²-x)-(4x-4)

                  =x(x-1)-4(x-4)

                  =(x-1)(x-4)

\(a,x^2-5xy+2x-10y\)

\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)

\(=x\left(x-5y\right)+2\left(x-5y\right)\)

\(=\left(x-5y\right)\left(x+2\right)\)

\(b,x^2-5x+4\)

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

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

\(=\left(x-1\right)\left(x-4\right)\)

Bài 1:

$5x+10=5(x+2)$

Bài 2:

Tại $x=8$ thì $x^2+4x+4=(x+2)^2=(8+2)^2=10^2=100$

Bài 3:

$x^2-6x+9=x^2-2.3.x+3^2=(x-3)^2$

Bài 4:

Diện tích mảnh đất là:

$(x+5)(x-5)=24$

$\Leftrightarrow x^2-25=24$

$\Leftrightarrow x^2=49$

$\Rightarrow x=7$ (do $x>5$)

Chiều dài mảnh đất là: $x+5=7+5=12$ (m)

\(8x^3+60x^2y+150xy^2+125y^3\)

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

\(=\left(2x+5y\right)\left(4x^2-10xy+25y^2\right)+30xy\left(2x+5y\right)\)

\(=\left(2x+5y\right)\left(4x^2-10xy+25y^2+30xy\right)\)

\(=\left(2x+5y\right)\left(4x^2+20xy+25y^2\right)\)

\(=\left(2x+5y\right)\left(2x+5y\right)^2=\left(2x+5y\right)^3\)

C2 : Áp dụng luôn HĐT ( x + y ) ³

\(8x^3+60x^2y+150xy^2+125y^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.5y+3.2x.\left(5y\right)^2+\left(5y\right)^3\)

\(=\left(2x+5y\right)^2\)

=.= hok tốt!!

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)

\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)

\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)

\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)

\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)

\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)