\(9a^2b^2-b^4+6b^3-9b^2\\ =b^2\left(9a^2-b^2+6b-9\right)\\ =b^2\left[9a^2-\left(b-3\right)^2\right]\\ =b^2\left(3a-b+3\right)\left(3a+b-3\right)\)

\(a^6+a^4+a^2b^2+b^4-b^6\\ =a^6-b^6+a^4+a^2b^2+b^4\\ =\left(a^6-b^6\right)+\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2\right)^3-\left(b^2\right)^3\right]+\left(a^4+a^2b^2+b^4\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^2+a^2b^2+b^4\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left(a^2-b^2+1\right)\left(a^4+2a^2b^2+b^4-a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left[\left(a^2+b^2\right)^2-\left(ab\right)^2\right]\\ =\left(a^2-b^2+1\right)\left(a^2+b^2-ab\right)\left(a^2+b^2+ab\right)\)

a6, a4 là số mũ hay hệ số vậy bn

Đặt \(A=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4\)

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

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

Áp dụng hàng đẳng thức \(\left(a^2-b^2+c^2\right)=a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ca\right)^2\):

\(A=-\left[\left(a^2-b^2+c^2\right)^2-4\left(ca\right)^2\right]\)

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

2222222222222a+257222222222222222222222222222222222222222222222222222222222222222222222222222222222222222a=?

a) \(25a^2-1=\left(5a-1\right)\left(5a+1\right)\)

b) \(a^2-9=\left(a-3\right)\left(a+3\right)\)

c) \(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)

d) \(\dfrac{9}{4}a^4-\dfrac{16}{25}=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\)

e) \(\left(2a+b\right)^2-a^2=\left(2a+b-a\right)\left(2a+b+a\right)=\left(a+b\right)\left(3a+b\right)\)

f) \(16\left(x-1\right)^2-25\left(x+y\right)^2=\left(4x-4-5x-5y\right)\left(4x-4+5x+5y\right)=\left(-x-4-5y\right)\left(9x+5y-4\right)\)

a/ $25x^2-1\\=(5x)^2-1^2\\=(5x-1)(5x+1)$

b/ $a^2-9\\=a^2-3^2\\=(a-3)(a+3)$

c/ $\dfrac{1}{4}a^2-\dfrac{9}{25}\\=\left(\dfrac{1}{2}a\right)^2-\left(\dfrac{3}{5}\right)^2\\=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)$

d/ $\dfrac{9}{4}a^4-\dfrac{16}{25}\\=\left(\dfrac{3}{2}a^2\right)^2-\left(\dfrac{4}{5}\right)^2\\=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left[\left(\sqrt{\dfrac 3 2}a\right)^2-\left(\dfrac{2\sqrt 5}{5}\right)^2\right]\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left(\sqrt{\dfrac 3 2}a-\dfrac{2\sqrt 5}{5}\right)\left(\sqrt{\dfrac 3 2}a+\dfrac{2\sqrt 5}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)$

e/ $(2a+b)^2-a^2\\=(2a+b-a)(2a+b+a)\\=(a+b)(3a+b)$

f/ $16(x-1)^2-25(x+y)^2\\=[4(x-1)]^2-[5(x-y)]^2\\=[4(x-1)-5(x-y)][4(x-1)+5(x-y)]\\=[4x-4-5x+5y][4x-4+5x-5y]\\=(-x+5y-4)(9x-5y-4)$

1) \(x\sqrt{y}+y\sqrt{x}=\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\)

2) \(9-6\sqrt{a}+a=\left(\sqrt{a}-3\right)^2\)

3) \(a+2\sqrt{ab}+b=\left(\sqrt{a}+\sqrt{b}\right)^2\)

4) \(x-y+\sqrt{x}+\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+1\right)\)

5) \(a+2\sqrt{ab}+b-1=\left(\sqrt{a}+\sqrt{b}\right)^2-1=\left(\sqrt{a}+\sqrt{b}-1\right)\left(\sqrt{a}+\sqrt{b}+1\right)\)

1) \(x\sqrt{y}+y\sqrt{x}=\sqrt{x}\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\)

2) \(9-6\sqrt{a}+a=\left(3-\sqrt{a}\right)^2\)

3) \(a+2\sqrt{ab}+b=\left(\sqrt{a}+\sqrt{b}\right)^2\)

4) \(x-y+\sqrt{x}+\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+1\right)\)

5) \(a+2\sqrt{ab}+b-1=\left(\sqrt{a}+\sqrt{b}\right)^2-1^2=\left(\sqrt{a}+\sqrt{b}-1\right)\left(\sqrt{a}+\sqrt{b}+1\right)\)

\(1,=\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\\ 2,=\left(\sqrt{a}-3\right)^2\\ 3,=\left(\sqrt{a}+\sqrt{b}\right)^2\\ 4,=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)\\ 5,=\left(\sqrt{a}+\sqrt{b}\right)^2-1=\left(\sqrt{a}+\sqrt{b}-1\right)\left(\sqrt{a}+\sqrt{b}+1\right)\)

=(c-b-a)(c-b+a)(c+b-a)(c+b+a)

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

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

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

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

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

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

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

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)