a, \(5x^2y+10xy=5xy\left(x+2\right)\)

b, \(x^2-2xy+y^2-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)

c, \(x^3-8+2x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)

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

d, \(x^4+x^2y^2+y^4\):< 

a, 5xy(  x + 2)

b, ( x - y -5 )(x-y+5)

c,( x-2)( x+ 2)²

bđt <=> \(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)

\(=\frac{bc}{a^2\left(b+c\right)}+\frac{ac}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\)

\(=\frac{1}{a^2\left(\frac{1}{c}+\frac{1}{b}\right)}+\frac{1}{b^2\left(\frac{1}{a}+\frac{1}{c}\right)}+\frac{1}{c^2\left(\frac{1}{b}+\frac{1}{a}\right)}\)

\(=\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{a}}\)(1)

Đặt 1/a = x ; 1/b = y ; 1/c = z

(1) được viết lại thành \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\)

Theo Bunyakovsky dạng phân thức ta có :

\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)

Lại có \(x+y+z\ge3\sqrt{xyz}=3\sqrt{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}=3\sqrt{\frac{1}{abc}}=3\)( Cauchy )

=> \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\ge\frac{x+y+z}{2}\ge\frac{3}{2}\)

Hay \(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\ge\frac{3}{2}\)( đpcm )

Đẳng thức xảy ra <=> x = y = z = 1 

hay a = b = c = 1

Ta có : a³ + b³ + c³ = 3abc

=> (a + b)(a² - ab + b²) + c³ - 3abc = 0

=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

=> [(a + b)³ + c³] - [(3ab(a + b) + 3abc] = 0

=> (a + b + c)(a² + b² + 2ab - ac - bc + c²) - 3ab(a + b + c) = 0

=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

=> a² + b² + c² - ab- ac - bc = 0

=> 2(a² + b² + c² - ab- ac - bc) = 0

=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

=> (a² - 2ab + b²) + (b² - 2bc + c²) + (a² - 2ac + c²) = 0

=> (a - b)² + (b - c)² + (a - c)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)

Khi đó M = \(\frac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}=\frac{3.c^{2020}}{\left(3c\right)^{2020}}+\frac{3c^{2020}}{3^{2020}.c^{2020}}=\frac{1}{3^{2019}}\)

( 2x + 5 )( 2x - 5 ) - ( 4x⁵ - 2x⁴ ) : ( -x³ ) = 15

<=> 4x² - 25 - [ ( 4x⁵ : ( -x³ ) - ( 2x⁴ : ( -x³ ) ] = 15

<=> 4x² - 25 - ( -4x² + 2x ) = 15

<=> 4x² - 25 + 4x² - 2x - 15 = 0

<=> 8x² - 2x - 40 = 0

Δ = b² - 4ac = (-2)² - 4.8.(-40) = 1284

Δ > 0 nên phương trình có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\text{Δ}}}{2a}=\frac{2+\sqrt{1284}}{16}=\frac{1+\sqrt{321}}{8}\\x_2=\frac{-b-\sqrt{\text{Δ}}}{2a}=\frac{2-\sqrt{1284}}{16}=\frac{1-\sqrt{321}}{8}\end{cases}}\)

Vậy ...

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

\(\Leftrightarrow\left(x+3\right)\left[x\left(x-1\right)-x^2\right]=-4\)

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

\(\Leftrightarrow-x\left(x+3\right)=-4\)

\(\Leftrightarrow-x^2-3x=-4\)

\(\Leftrightarrow x^2+3x-4=0\)

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

\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

vậy........

x( x - 1 )( x + 3 ) - x²( x + 3 ) = -4

⇔ ( x + 3 )[ x( x - 1 ) - x² ] + 4 = 0

⇔ ( x + 3 )( x² - x - x² ) + 4 = 0

⇔ ( x + 3 ).(-x) + 4 = 0

⇔ -x² - 3x + 4 = 0

⇔ -( x² + 3x - 4 ) = 0

⇔ -( x² - x + 4x - 4 ) = 0

⇔ -[ x( x - 1 ) + 4( x - 1 ) ] = 0

⇔ -( x - 1 )( x + 4 ) = 0

⇔ x - 1 = 0 hoặc x + 4 = 0

⇔ x = 1 hoặc x = -4