vì:

(a-b)-(a+b)

= a-b-a+b

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

=2b + 0

=2b

Ta có : 

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

\(=a-b-a+b\)

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

\(=b+b+0\)

\(=2b+0=2b\)

Ta có:

(x² + y²)² - 4x²y² = (x² + y²)² - (2xy)² = (x² + y² - 2xy)(x² + y² + 2xy) = (x - y)²(x + y)²

(x^2+y^2)^2-4x^2y^2

=x^4+2x^2y^2+y^4-4x^2y^2

=x^4-2x^2y^2+^4

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

hok tốt he

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

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

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

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

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

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

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

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

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

 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