1.a) (3x+1)²-4(x-2)²= (3x+1)²-[2(x-2)]²=[(3x+1)-2(x-2)][(3x+1)+2(x-2)]=(x+3)(5x-1)

b) (a²+b²-5)²-4(ab+2)²= (a²+b²-5)²-[2(ab+2)]² = (a²+b²-5-2ab-4)(a²+b²-5+2ab+4)=[(a-b)²-9][(a+b)²-1]

2. 3x²+9x-30=3x²-6x+15x-30=3x(x-2)+15(x-2)=3(x+5)(x-2)

b. x³-5x²-14x=x³+2x²-7x²-14x=x²(x+2)-7x(x+2)=(x²-7x)(x+2)

a) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)

\(=\left(3x+1\right)^2-\left[2.\left(x-2\right)\right]^2\)

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

\(=\left[3x+1-2x+4\right].\left[3x+1+2x-4\right]\)

\(=\left(x+5\right)\left(5x-3\right)\)

b) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5\right)^2-\left[2.\left(ab+2\right)\right]^2\)

\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right].\left[\left(a+b\right)^2-1\right]\)

\(=\left[\left(a-b-3\right)\left(a-b+3\right)\right].\left[\left(a+b-1\right)\left(a+b+1\right)\right]\)

a) \(3x^2+9x-30\)

\(=3\left(x^2+3x-10\right)\)

\(=3\left(x^2-2x+5x-10\right)\)

\(=3.\left[x\left(x-2\right)+5.\left(x-2\right)\right]\)

\(=3.\left[\left(x+5\right)\left(x-2\right)\right]\)

b) \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2+2x-7x-14\right)\)

\(=x.\left[x\left(x+2\right)-7.\left(x+2\right)\right]\)

\(=x.\left[\left(x-7\right)\left(x+2\right)\right]\)

a) 4x²-1=(2x)²-1²=(2x-1)(2x+1)

b)25x²-0.09=(5x)²-\(\left(\frac{3}{10}\right)^2\)=\(\left(5x-\frac{3}{10}\right)\left(5x+\frac{3}{10}\right)\)

c)9x²-14=(3x)²-\(\sqrt{14}^2\)=(3x-\(\sqrt{14}\))(3x+\(\sqrt{14}\))

d) (x-y)²-4=(x-y)²-2²=(x-y-2)(x-y+2)

e) 9-(x-y)²=3³-(x-y)²=(3-x+y)(3+x-y)

f)(x²+4)²-16x²=(x²+4)²-(4x)²=(x²+4+4x)(x²+4-4x)

Chúc hok tốt!!!

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

\(b,\)\(25x^2-0,09=\left(5x\right)^2-0,3^2=\left(5x-0,3\right)\left(5x+0,3\right)\)

\(c,\)\(9x^2-14=\left(3x\right)^2-\left(\sqrt{14}\right)^2=\left(3x-\sqrt{14}\right)\left(3x+\sqrt{14}\right)\)

\(d,\)\(\left(x-y\right)^2-4=\left(x-y\right)^2-2^2=\left(x-y+2\right)\left(x-y-2\right)\)

\(e,\)\(9-\left(x-y\right)^2=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)

\(f,\)\(\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)^2-\left(4x\right)^2=\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)

                                             \(=\left(x^2-2.2x+2^2\right)\left(x^2+2.2x+2^2\right)\)

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

a) 3x² - 5x - 3y² + 5y

= 3(x²- y²) -5(x-y)

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

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

b) 49 - x²+2xy-y²

= 7² - (x-y)²

=(7-x+y)(7+x-y)

a) \(3x^2-5x-3y^2+5y\)

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

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

\(=3\left[\left(x-y\right).\left(x+y\right)\right]-5\left(x-y\right)\)

\(3\left(x-y\right).\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right).\left[3.\left(x+y\right)-5\right]\)

\(=\left(x-y\right)\left(3x+3y-5\right)\)

b) \(49-x^2+2xy-y^2\)

\(=7^2-x^2+2xy-y^2\)

\(=7^2-\left(x^2-2xy+y^2\right)\)

\(=7-\left(x-y\right)^2\)

\(=\sqrt{7}^2-\left(x-y\right)^2\)

\(=\left[7-\left(x-y\right).-7+\left(x-y\right)\right]\)

\(=\left(7-x+y\right).\left(-7+x-y\right)\)

<=>x³-3x²+3x-1+1+3x²-12x=0

<=> x³-9x=0

<=>x(x²-9)=0

<=> x(x-3)(x+3)=0

<=>\(\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

chúc hok tốt!!!

16x³+27

=(\(\sqrt{16}\)x)³+3³

=\(\left(\sqrt{16}x+3\right)\left[\left(\sqrt{16}x\right)^2-3.\sqrt{16}x+9\right]\)

,Ta có E là trung điểm của AB nên AE=EB=18

               AE^2+AD^2=DE^2
nên 1872=DE^2 nên DE=30
lại có t/g AED=T/G BEG(G.C.G) nên EG=DE nên DG=2DE=60
ta có: t/g AFE đồng dạng t/g CFD(g.g)
nên DF/FE=DC/AE→ DF/(DF+EF)=DC/(AE+CD)
NÊN DF/DE=AE/54→DF/30=36/54 NÊN DF=20

Gọi 3 số tự nhiên liên tiếp đó là: \(n-1;n;n+1\left(n\in N\right)\)

Theo đề bài, ta có: \(n\left(n+1\right)-n\left(n-1\right)=12\)

                 \(\Leftrightarrow n^2+n-n^2+n=12\)

                  \(\Leftrightarrow2n=12\Leftrightarrow n=6\) (thỏa đk)

Vậy 3 số tự nhiên liên tiếp cần tìm là 5;6;7

Có: ab(x²+y²) - xy(a²+b²)

= abx² + aby² - xya² - xyb²

= ax(bx-ay) - by(bx-ay)

=(ax-by)(bx-ay)

\(ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)\)

\(=abx^2+aby^2-xya^2-xyb^2\)