Bài 1 :

Ta có : a + b + c = 0

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

Ta có : a³ + b³ + c³ 

= ( a³ + b³ ) + c³

= ( a + b )³ - 3ab . ( a + b ) + c³ ( 1 )

Thay a + b = - c vào ( 1 ) , ta được :

- c³ - 3ab . ( - c ) + c³ = 3ab

Hay a³ + b³ + c³ = 3ab ( đpcm )

\(C=\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right):\left(1-\frac{2x}{x^2+1}\right)\)

\(C=\left(\frac{1}{x-1}-\frac{2x}{x^2\left(x-1\right)+x-1}\right):\left(\frac{x^2+1-2x}{x^2+1}\right)\)

\(C=\left(\frac{1}{x-1}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right):\left[\frac{\left(x-1\right)^2}{x^2+1}\right]\)

\(C=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}.\frac{x^2+1}{\left(x-1\right)^2}\)

\(C=\frac{1}{x-1}\)

a ) ( x² + 2x + 5 )( x² + 2x + 3 ) - 8

= ( x² + 2x + 5 )[ ( x² + 2x + 5 ) - 2 ] - 8

= ( x² + 2x + 5 )² - 2 . ( x² + 2x + 5 ) + 1 - 9

= ( x² + 2x + 5 - 1 )² - 9

= ( x² + 2x + 4 )² - 3³

= ( x² + 2x + 4 - 3 )( x² + 2x + 4 + 3 )

= ( x² + 2x + 1 )( x² + 2x + 7 )

b ) ( x² + 2x )( x² + 2x - 2 ) - 3

= ( x² + 2x )[ ( x² + 2x ) - 2 ] - 3 

= ( x² + 2x )² - 2 . ( x² + 2x ) + 1 - 4 

= ( x² + 2x - 1 )² - 2²

= ( x² + 2x - 1 - 2 )( x² + 2x - 1 + 2 )

= ( x² + 2x - 3 )( x² + 2x + 1 )

= ( x² + 2x - 3 )( x + 1 )²

trả lời :

• \(\left(x^2+2x+5\right)\left(x^2+2x+3\right)\)

Đặt: \(x^2+2x+5=t\Rightarrow x^2+2x+3=t+2\),ta có:

\(t\left(t+2\right)-8\)

\(=t^2+2t-8\)

\(=t^2+4t-2t-8\)

\(=t\left(t+4\right)-2\left(t+4\right)\)

\(=\left(t+4\right)\left(t-2\right)\)

Thay vào cách đặt , ta có:

\(\left(x^2+2x+5+4\right)\left(x^2+2x+5-2\right)\)

\(=\left(x^2+2x+9\right)\left(x^2+2x+3\right)\)

\(=\left(x^2+2x+9\right)\left(x^2+3x-x+3\right)\)

\(=\left(x^2+2x+9\right)\left(x+3\right)\left(x-1\right)\)

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

Đặt : \(x^2+2x=t\Rightarrow\left(x^2+2x-2\right)=t-2\),ta có:

\(t\left(t-2\right)-3\)

\(=t^2-2t-3\)

\(=t^2-3t+t-3\)

\(=t\left(t-3\right)+\left(t-3\right)\)

\(=\left(t-3\right)\left(t+1\right)\)

Thay vào cách đặt, ta có:

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

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

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

#hok tốt #

x binh + y binh + z binh = 1

Bạn giải chi tiết đc k?

Ta có : ( x² - 1 )( x² + 4x + 3 ) = 0

\(\Leftrightarrow\)( x - 1 )( x + 1 )[ ( x² + 4x + 4 ) - 1 ] = 0

\(\Rightarrow\)( x - 1 )( x + 1 )[ ( x + 2 )² - 1² ] = 0

\(\Rightarrow\)( x - 1 )( x + 1 )( x + 2 - 1 )( x + 2 + 1 ) = 0

\(\Rightarrow\)( x - 1 )( x + 1 )( x + 1 )( x + 3 ) = 0

\(\Rightarrow\)( x - 1 )( x + 1 )²( x + 3 ) = 0

\(\Rightarrow\)x - 1 = 0 hoặc ( x + 1 )² = 0 hoặc x + 3 = 0

\(\Rightarrow\)x = 1 hoặc x = - 1 hoặc x = - 3 

Vậy : x = 1 hoặc x = - 1 hoặc x = - 3

(x^2-1)(x^2+4x+3)=0

(x-1)(x+1)(x^2+3x+x+3)=0

(x-1)(x+1)[ x(x+1)+3(x+1)]=0

(x-1)(x+1)(x+3)(x+1)=0

x-1=0 hoặc x+1=0 hoặc x+3=0 

=> x=1;-1;-3

\(\frac{x+11}{115}+\frac{x+22}{104}=\frac{x+33}{93}+\frac{x+44}{82}\)

\(\Leftrightarrow\frac{1+\left(x+11\right)}{115}+\frac{1+\left(x+22\right)}{104}=\frac{1+\left(x+33\right)}{93}+\frac{1+\left(x+44\right)}{82}\)

\(\Leftrightarrow\frac{x+126}{115}+\frac{x+126}{104}=\frac{x+126}{93}+\frac{x+126}{82}\)

\(\Leftrightarrow\left(x+126\right).\left(\frac{1}{115}+\frac{1}{104}+\frac{1}{93}+\frac{1}{82}\right)=0\)

\(\Leftrightarrow x+126=6\Leftrightarrow x=-126\)vì\(\frac{1}{115}+\frac{1}{104}+\frac{1}{93}+\frac{1}{82}\ne0\)

vậy x=-126

\(x^8+x^4+1\)

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

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