a ) a ³ (b - c) + b ³ (c - a)+ c ³ (a - b)

=a³(b-c)-b³[(b-c)+(a-b)]+c³(a-b)

=a³(b-c)-b³(b-c)-b³(a-b)+c³(a-b)

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

=(b-c)(a-b)(a²+2b²+c²+ab+bc)

\(a,x^2-25-x-5=0\)

\(x^2-x-30=0\)

\(x^2+5x-6x-30=0\)

\(x\cdot\left(x+5\right)-6\cdot\left(x+5\right)=0\)

\(\left(x+5\right)\cdot\left(x-6\right)=0\)

\(\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}}\)

b) \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Leftrightarrow\left(10x^2+9x\right)-\left(10x^2+13x-3\right)=8\)

\(\Leftrightarrow-4x+3=8\)

\(\Leftrightarrow-4x=5\Leftrightarrow x=\frac{-5}{4}\)

Ta có: \(P=-x^2+2x+5\)

\(=-x^2+2x-1+6\)

\(=-\left(x-1\right)^2-6\)

Vì \(-\left(x-1\right)^2\le0;\forall x\)

\(\Rightarrow-\left(x-1\right)^2-6\le0-6;\forall x\)

Hay\(P\le-6;\forall x\)

Dấu "="xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)

                        \(\Leftrightarrow x=1\)

Vậy MAX P=-6 khi x=1

\(x^2-x+27=0\)

Ta có: \(\Delta=1-4.27=-107< 0\)

Vậy pt vô nghiệm

ta có:

a) (x² - 3x + xy - 3y) : (x + y)

= [x(x - 3) + y(x - 3)] : (x + y)

= (x + y)(x - 3) : (x + y)

= x - 3

b) (x² - y² + 6x + 9) : (x + y + 3)

= [(x² + 6x + 9) - y²] : (x + y + 3)

= [(x + 3)² - y²] : (x + y + 3)

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

= x - y + 3

Cậu đặt phép chia, nó sẽ ra : 

2x³ -x² -x +1 = (x-2)(2x² + 3x + 5) dư 11

=> 2x³ -x² -x +1 \(⋮\)x-2 \(\Leftrightarrow\)x-2=11

                                      \(\Leftrightarrow\)x=13

\(2x^2\text{ chẵn};2011\text{ lẻ}\Rightarrow y^2\text{ lẻ}\Rightarrow y^2\text{ chia 8 dư 1}\Rightarrow2x^2\text{ chia 8 dư 4}\Leftrightarrow x^2\text{ chia 4 dư 2}\)

bđt \(\Leftrightarrow\)\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\ge3a^3b+3b^3c+3c^3a\)

Có: \(a^4+a^2b^2\ge2a^3b\) tương tự với b, c, do đó cần cm: \(a^2b^2+b^2c^2+c^2a^2\ge a^3b+b^3c+c^3a\)

\(\Leftrightarrow\)\(a^2b\left(b-a\right)+b^2c\left(c-b\right)+c^2a\left(a-c\right)\ge0\) (1) 

Do a,b,c vai trò như nhau nên giả sử \(0\le a\le b\le c\) ta có: 

\(c^2a\left(a-c\right)=c.c.a\left(a-c\right)\ge b.a.a\left(a-c\right)=a^2b\left(a-c\right)\)

\(\Rightarrow\)\(VT_{\left(1\right)}\ge a^2b\left(b-a\right)+b^2c\left(c-b\right)+a^2b\left(a-c\right)=a^2b\left(b-a+a-c\right)+b^2c\left(c-b\right)\)

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

Mà \(0\le a\le b\le c\) nên \(\hept{\begin{cases}b-c\le0\\a^2-bc\le0\end{cases}}\)\(\Rightarrow\)\(VT_{\left(1\right)}\ge b\left(b-c\right)\left(a^2-bc\right)\ge0\)

Phùng Minh Quân vai trò của a,b,c không như nhau nhé