X³ +  X² + 4 

X⁵ + 4 

x³ + x² + 4

= x³ - 4x + x² + 4x + 4

= ( x³ - 4x ) + ( x² + 4x + 4 )

= x( x² - 4 ) + ( x + 2 )²

= x( x - 2 )( x + 2 ) + ( x + 2 )²

= ( x + 2 )[ x( x - 2 ) + ( x + 2 ) ]

= ( x + 2 )( x² - 2x + x + 2 )

= ( x + 2 )( x² - x + 2 )

tui nè

cậu có nick lazi ko ???

Mk trao đổi

a) ( 3x - 1 )² - 16 = ( 3x - 1 )² - 4² = ( 3x - 1 - 4 )( 3x - 1 + 4 ) = ( 3x - 5 )( 3x + 3 ) = 3( 3x - 5 )( x + 1 )

b) ( 5x - 4 )² - 49x² = ( 5x - 4 )² - ( 7x )² = ( 5x - 4 - 7x )( 5x - 4 + 7x ) = ( -2x - 4 )( 12x - 4 ) = -2( x + 2 ).4( 3x - 1 ) = -8( x + 2 )( 3x - 1 )

c) ( 2x + 5 )² - ( x - 9 )² = [ ( 2x + 5 ) - ( x - 9 ) ][ ( 2x + 5 ) + ( x - 9 ) ] = ( 2x + 5 - x + 9 )( 2x + 5 + x - 9 ) = ( x + 14 )( 3x - 4 )

d) ( 3x + 1 )² - 4( x - 2 )² = ( 3x + 1 )² - 2²( x - 2 )² = ( 3x + 1 )² - [ 2( x - 2 ) ]² = ( 3x + 1 )² - ( 2x - 4 )² = [ ( 3x + 1 ) - ( 2x - 4 ) ][ ( 3x + 1 ) + ( 2x - 4 ) ] = ( 3x + 1 - 2x + 4 )( 3x + 1 + 2x - 4 ) = ( x + 5 )( 5x - 3 )

e) 9( 2x + 3 )² - 4( x + 1 )² = 3²( 2x + 3 )² - 2²( x + 1 )² = [ 3( 2x + 3 ) ]² - [ 2( x + 1 ) ]² = ( 6x + 9 )² - ( 2x + 2 )² = [ ( 6x + 9 ) - ( 2x + 2 ) ][ ( 6x + 9 ) + ( 2x + 2 ) ] = ( 6x + 9 - 2x - 2 )( 6x + 9 + 2x + 2 ) = ( 4x + 7 )( 8x + 11 )

f) 4b²c² - ( b² + c² - a² )² = ( 2bc )² - ( b² + c² - a² )² = [ 2bc - ( b² + c² - a² ) ][ 2bc + ( b² + c² - a² ] = ( 2bc - b² - c² + a² )( 2bc + b²+ c² - a² ) = [ a² - ( b² - 2bc + c² ) ][ ( b² + 2bc + c² ) - a² ] = [ a² - ( b - c )² ][ ( b + c )² - a² ] = ( a - b + c )( a + b - c )( b + c - a )( b + c + a )

g) ( ax + by )² - ( ay + bx )² 

= [ ( ax + by ) - ( ay + bx ) ][ ( ax + by ) + ( ay + bx ) ]

= ( ax + by - ay - bx )( ax + by + ay + bx )

= [ a( x - y ) - b( x - y ) ][ a( x + y ) + b( x + y ) ]

= ( a - b )( x - y )( x + y )( a + b )

h) ( a² + b² - 5 )² - 4( ab + 2 )² 

= ( a² + b² - 5 )² - 2²( ab + 2 )² 

= ( a² + b² - 5 )² - [ 2( ab + 2 ) ]² 

= ( a² + b² - 5 )² - ( 2ab + 4 )² 

= [ ( a² + b² - 5 ) - ( 2ab + 4 ) ][ ( a² + b² - 5 ) + ( 2ab + 4 ) ]

= ( a² + b² - 5 - 2ab - 4 )( a² + b² - 5 + 2ab + 4 )

= [ ( a² - 2ab + b² ) - 9 ][ ( a² + 2ab + b² ) - 1 ]

= [ ( a - b )² - 3² ][ ( a + b )² - 1² ]

= ( a - b - 3 )( a - b + 3 )( a + b - 1 )( a + b + 1 )

i) ( 4x² - 3x - 18 )² - ( 4x² + 3x )²

= [ ( 4x² - 3x - 18 ) - ( 4x² + 3x ) ][ ( 4x² - 3x - 18 ) + ( 4x² + 3x ) ]

= ( 4x² - 3x - 18 - 4x² - 3x )( 4x² - 3x - 18 + 4x² + 3x )

= ( -6x - 18 )( 8x² - 18 )

= -6( x + 3 ).2( 4x² - 9 )

= -12( x + 3 )( 2x - 3 )( 2x + 3 )

k) 9( x + y - 1 )² - 4( 2x + 3y + 1 )²

= 3²( x + y - 1 )² - 2²( 2x + 3y + 1 )²

= [ 3( x + y - 1 ) ]² - [ 2( 2x + 3y + 1 ) ]²

= ( 3x + 3y - 3 )² - ( 4x + 6y + 2 )²

= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]

= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )

= ( -x - 3y - 5 )( 7x + 9y - 1 )

l) -4x² + 12xy - 9y² + 25

= 25 - ( 4x² - 12xy + 9y² )

= 5² - ( 2x - 3y )²

= ( 5 - 2x + 3y )( 5 + 2x - 3y )

m) x² - 2xy + y² - 4m² + 4mn - n²

= ( x² - 2xy + y² ) - ( 4m² - 4mn + n² )

= ( x - y )² - ( 2m - n )²

= ( x - y - 2m + n )( x - y + 2m - n )

P = x³ + y³ - x² - y² + 3xy( x + y ) - 2xy + 3( x + y ) + 10

= ( x³ + y³ ) - ( x² + 2xy + y² ) + 3xy( x + y ) + 3.5 + 10

= ( x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² ) - ( x + y )² + 3xy( x + y ) + 15 + 10

= [ ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² ) ] - 5² + 3xy( x + y ) + 25

= ( x + y )³ - 3xy( x + y ) - 25 + 3xy( x + y ) + 25

= 5³ = 125

8x³ - 4x² + 2x - 1 = 0

<=> ( 8x³ - 4x² ) + ( 2x - 1 ) = 0

<=> 4x²( 2x - 1 ) + ( 2x - 1 ) = 0

<=> ( 2x - 1 )( 4x² + 1 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\4x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\) ( do 4x² + 1 ≥ 1 > 0 ∀ x )

\(8x^3-4x^2+2x-1=0\)

\(\left(8x^3-4x^2\right)+\left(2x-1\right)=0\)

\(4x^2\left(2x-1\right)+\left(2x-1\right)=0\)

\(\left(2x-1\right)\left(4x^2+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\4x^2+1=0\end{cases}\Rightarrow x=\frac{1}{2}}\)

Đề bài??

đề bải là j bạn quên rồi kìa

Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

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

\(\Leftrightarrow ab+bc+ca=0\)

\(\Rightarrow\hept{\begin{cases}ab=-bc-ca\\bc=-ca-ab\\ca=-ab-bc\end{cases}}\)

Thay vào ta được: \(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2+bc-ca-ab}=\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)

Tương tự: \(\frac{b^2}{b^2+2ca}=\frac{b^2}{\left(b-a\right)\left(b-c\right)}\) ; \(\frac{c^2}{c^2+2ab}=\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(\Rightarrow P=-\left[\frac{a^2}{\left(a-b\right)\left(c-a\right)}+\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\right]\)

\(=-\left[\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right]\)

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

\(=\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow ab+ac+bc=0\)

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

Tương tự: \(\frac{b^2}{b^2+2ac}=\frac{b^2}{\left(b-a\right)\left(b-c\right)};\frac{c^2}{c^2+2ac}=\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

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

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

Dễ mà :

\(4x^2+6x\)

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

Học tốt

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+z^2x+zx^2+3xyz-xyz=0\)

\(\Leftrightarrow\left(x^2y+xy^2\right)+\left(yz^2+z^2x\right)+\left(zx^2+2xyz+y^2z\right)=0\)

\(\Leftrightarrow xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(xy+z^2+yz+zx\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=> x = -y hoặc y = -z hoặc z = -x

Không mất tổng quát giả sử x = -y, khi đó:

\(\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=-\frac{1}{y^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{z^{2015}}\)

\(\frac{1}{x^{2015}+y^{2015}+z^{2015}}=\frac{1}{-y^{2015}+y^{2015}+z^{2015}}=\frac{1}{z^{2015}}\)

\(\Rightarrow\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{x^{2015}+y^{2015}+z^{2015}}\)