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) \(x^3-5x^2-14x\)

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

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

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

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

dấu ^ là mũ nha mn

giups mình với nha

a) Đặt a + b = x ; a - b = y. Khi đó:
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(\Leftrightarrow x^3-y^3\)
\(\Leftrightarrow\left[x-y\right]\left[x^2+xy+y^2\right]\)
Thế lại vào ta có:
\(\Leftrightarrow\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(\Leftrightarrow\left[\left(a-a\right)+\left(b+b\right)\right]\left[\left(a^2+b^2+2ab\right)+\left(a^2-b^2\right)+\left(a^2+b^2-2ab\right)\right]\)
\(\Leftrightarrow2b\left[\left(a^2+a^2+a^2\right)+\left(b^2-b^2+b^2\right)+\left(2ab-2ab\right)\right]\)
\(\Leftrightarrow2b\left[3a^2+b^2\right]\)

Mik làm tuỳ theo mình piết thôi nhé

a)   ( a + b )³- ( a - b )³= a³ + b³ - a³ - b³ = a³ - a³ + b³ - b³ = 0

b) tương tự như ở trên!!! Hơi khác một tí!!!

c)   ( 6x - 1 )² - ( 3x + 2 ) = ..........

\(\left(a+b\right)^3-\left(a-b\right)^3\)

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

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

\(\left(a+b\right)^3+\left(a-b\right)^3\)

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

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

a) (a+b)³ -(a-b)³ = a³ + 3a²b + 3ab² +b³ - a³ + 3a²b - 3ab² +b³

                       = 2a³ + 6a²b  + 2b³

\(\left(6x-1\right)^2-\left(3x+2\right)\)

\(=36x^2-12x+1-3x-2\)

\(=36x^2-15x-1\)

bn ktra lại đề nhé

hk tốt

\(=\left(4-a+b\right)\left(4+a-b\right)\)

a) \(\left(2x+5\right)^2\)\(-\left(x-9\right)^2\)

=\(\left(2x+5+x-9\right).\left(2x+5-x+9\right)\)

=\(\left(3x-4\right).\left(x+14\right)\)