(x²+x+1)(x²-x+1)(x²-1)

= x^6-1

nha bạn chúc bạn học tốt nha 

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

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

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

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

\(=x^6-1\)

= ( 2x + 3 - 2x - 5 )² = (-2)² = 4

(2x+3)²-2(2x+3)(2x+5)+(2x+5)²

<=> [ (2x+3)-(2x+5) ]²

<=> (-2)²

<=> 4

 #hoctot# 

lm hết đống này á

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

\(=-2\left(x^2-\left|x\right|\right)-5\)

\(=-2\left(\left|x\right|^2-2\left|x\right|\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}-5\)

\(=-2\left(\left|x\right|-\frac{1}{2}\right)^2-\frac{19}{5}\)

Mà: \(\left(\left|x\right|-\frac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow-2\left(\left|x\right|-\frac{1}{2}\right)^2\le0\)

\(\Leftrightarrow-2\left(\left|x\right|-\frac{1}{2}\right)^2-\frac{19}{5}\le\frac{-19}{5}\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|x\right|-\frac{1}{2}=0\Leftrightarrow\left|x\right|=\frac{1}{2}\Leftrightarrow x=\pm\frac{1}{2}\)

x^3y^6/8 - 27/y^3 

= -27x^(3y^6)/(8y^3)

e) \(\frac{x^3y^6}{8}-\frac{27}{y^3}=\left(\frac{xy^2}{2}\right)^3-\left(\frac{3}{y}\right)^3\)

\(=\left(\frac{xy^2}{2}-\frac{3}{y}\right)\left(\frac{x^2y^4}{4}+\frac{3xy}{2}+\frac{9}{y^2}\right)\)

\(27x^3+8y^3\)

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

\(=\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)

27x³ + 8y³

=(3x)³ +(2y)³

=(3x + 2y)[ (3x)² - 3x . 2y + (2y)² ]

= (3x+2y)(9x² -6xy + 4y²)

Các bạn giúp mình nhé mình đang cần sớm ạ thank you!
 

\(\left(2x^2y-xy\right)^3\)

\(=\left(2x^2y\right)^3-3.\left(2x^2y\right)^2.xy+3.2x^2y.\left(xy\right)^2+\left(xy\right)^3\)

\(=8x^6y^3-12x^5y^3+6x^4y^3-x^3y^3\)