a/ 

\(\left(5xy^2-11x^3y+6x^2y^2\right)\div x^2y\)

\(=xy\left(5y-11x^2+6xy\right)\div x^2y\)

\(=\left(5y-11x^2+6xy\right)\div x\)

\(=\frac{5y}{x}-\frac{11x^2}{x}+\frac{6xy}{x}\)

\(=\frac{5y}{x}-11x+6y\)

b/ \(\left[\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3\right]\div\left[-5\left(x+y\right)^3\right]\)

\(=\left(x+y\right)^3\left[\left(x+y\right)^2-2\left(x+y\right)+3\right]\div\left[-5\left(x+y\right)^3\right]\)

\(=\frac{\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)

Có gửi card không bạn ơi?

b)(x-2y)(x+2y)

=x^2-4y^2

c)(x-1)(x^2+x+1)

=x^3-1

Giup mik vs

a/

\(2x\left(x+5\right)\left(x-1\right)\)

\(=\left(2x^2+10x\right)\left(x-1\right)\)

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

\(=2x^3+8x^2-10x\)

b/

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

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

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

c/

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

\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)\)

\(=x^3-1^3\)

\(=x^3-1\)

(x-1)(x+2)=x+2

=>x-1=1

=>x=2

Ta có:

\(\left(x-1\right)\left(x+2\right)-x-2=0\)

\(\Leftrightarrow x^2+x-2-x-2=0\)

\(\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow x+2=0\)hoặc \(x-2=0\)

\(\Leftrightarrow x=-2\)hoặc \(x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{-2;2\right\}\)

Ta có:

       \(P=\left(x^3-2x^2+x-1\right)\left(5x^4-x\right)\)

\(\Leftrightarrow P=5x^7-x^4-10x^6+2x^3+5x^5-x^2-5x^4+x\)

\(\Leftrightarrow P=5x^7-10x^6+5x^5-6x^4+2x^3-x^2+x\)

Vậy hệ số của x⁴ trong đa thức P là: -6

a) \(x^3+6x^2+3x-10\)

\(=x^3-x^2+7x^2-7x+10x-10\)

\(=x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+7x+10\right)\)

\(=\left(x-1\right)\left(x^2+2x+5x+10\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x+5\right)\)

b)  \(x^3+3x^2-33x-35\)

\(=x^3-5x^2+8x^2-40x+7x-35\)

\(=x^2\left(x-5\right)+8x\left(x-5\right)+7\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+8x+7\right)\)

\(=\left(x-5\right)\left(x^2+x+7x+7\right)\)

\(=\left(x-5\right)\left(x+1\right)\left(x+7\right)\)

\(3x^2+3x-6\)

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

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

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

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

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

          \(=a^3+b^3=VT\)

b) \(VT=a^3+b^3+c^3-3abc\)

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

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

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

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

a.

Xét vế phải, ta có : \(\left(a+b\right)^3-3ab\left(a+b\right)\)= \(\left(a^3+3a^2b+3ab^2+b^3\right)-\left(3a^2b+3ab^2\right)\)

=\(a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)=\(a^3+b^3\)(đpcm)

b

Xét vế phải, ta có \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)= ..........

Ý b bạn nhân vế phải vào rồi rút gọn sẽ ra vế trái :)