\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)

\(C=ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b-a+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b\right)+bc\left(a-c\right)+ac\left(a-c\right)=b\left(a+b\right)\left(a-c\right)+c\left(a-c\right)\left(a+b\right)=\left(a+b\right)\left(c+c\right)\left(a-c\right)\)

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

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

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

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

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

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

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

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

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

MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

a(b³ - c³) + b(c³ - a³) + c(a³ - b³)

= a(b³ - c³ ) + b( c³ - b³ + b³ - a³) + c(a³ - b³)

= a(b³ - c³) + b(c³ - b³) + b(b³ - a³) + c(a³ - b³)

= a(b³ - c³) - b(b³ - c³) - [b(a³ - b³) - c(a³- b³)]

= (b³ - c³)(a - b) - (a³- b³)(b - c)

= (b - c)(b² + bc + c²)(a - b) - (a - b)(a² + ab + b²)(b - c)

= (b - c)(a - b)(b² + bc + c² - a² + ab - b²)

= (b - c)(a - b) [ (c²  - a²) + (bc - ab) ]

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

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

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

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

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

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

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

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

\(y^2-6y+9-z^2\)

\(=\left(y-3\right)^2-z^2\)

\(=\left(y-3-z\right)\left(y-3+z\right)\)

Bài 1:

c. \(a^3-a^2x-ay+xy\)

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

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

Bài 2:

a. \(x^2-6x=0\)

\(x\left(x-6\right)=0\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

b) \(x^3-2x2+x=0\)

\(x\left(x^2-3\right)=0\)

\(\Rightarrow\)\(\hept{\begin{cases}x=0\\x^2=3\end{cases}}\)

\(\Rightarrow\)\(\hept{\begin{cases}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)

 .\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)

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

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

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

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

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