\(P=\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

Ta có:Tử số: \(x^3-y^3+z^3+3xyz\)

     \(=\left(x+z\right)^3-3zx\left(x+z\right)-y^3+3xyz\)

    \(=\left(x+z-y\right)\left[\left(x+z\right)^2+y\left(x+z\right)+y^2\right]-3xz\left(x+z-y\right)\)

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

         Mẫu số: \(\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\)

   \(=x^2+2xy+y^2+y^2+2yz+z^2+z^2-2zx+x^2\)

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

Suy ra \(P=\frac{x-y+z}{2}\)

\(\frac{x^3-y^3}{2y}.\left[\frac{2y}{4-2y-2x+xy}+\frac{2xy+4y}{\left(x-y\right)\left(x^2-4\right)}\right]\)

\(=\frac{x^3-y^3}{2y}.\left[\frac{2y}{\left(x-2\right)\left(y-2\right)}+\frac{2y\left(x+2\right)}{\left(x-y\right)\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x^3-y^3\right).\left[\frac{x-y}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}+\frac{y-2}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}\right]\)

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

\(=\frac{x^2+xy+y^2}{y-2}\)

a) 4x² - 12x - 7 = 4x² - 12x + 9 - 16 = ( 4x² - 12x + 9 ) - 16 = ( 2x - 3 )² - 4² = ( 2x - 3 - 4 )( 2x - 3 + 4 ) = ( 2x - 7 )( 2x + 1 )

b) x² + 6x - 1 ( nghiệm vô tỉ )

c) x² - 4x + 2 ( nghiệm vô tỉ )

d) x² + 4x - 12 = x² - 2x + 6x - 12 = x( x - 2 ) + 6( x - 2 ) = ( x - 2 )( x + 6 )

e) x² - 4x - 32 = ( x² - 4x + 4 ) - 36 = ( x - 2 )² - 6² = ( x - 2 - 6 )( x - 2 + 6 ) = ( x - 8 )( x + 4 )

f) x² + 2x - 1 ( nghiệm vô tỉ )

g) x⁴ - 5x² = x²( x² - 5 )

h) Đặt t = x + 2y - 3

<=> t² - 4t + 4

= ( t - 2 )²

= ( x + 2y - 3 - 2 )²

= ( x + 2y - 5 )²

i) x³( x² + 1 )² - 49x

= x[ x²( x² + 1 )² - 49 ]

= x[ ( x³ + x )² - 7² ]

= x( x³ + x - 7 )( x³ + x + 7 )

k) x³ + y³ + z³ - 3xyz

= ( x³ + y³ ) + z³ - 3xyz

= ( x + y )³ - 3xy( x + y ) + z³ - 3xyz

= [ ( x + y )³ + z³ ] - [ 3xy( x + y ) + 3xyz ]

= ( x + y + z )[ ( x + y )² - ( x + y )z + z² ] - 3xy( x + y + z )

= ( x + y + z )( x² + 2xy + y² - xz - yz + z² - 3xy )

= ( x + y + z )( x² + y² + z² - xy - yz - xz )

\(4x^4-21x^2y^2+4y^2=\left(2x^2\right)^2-2.2x^2.2y^2+\left(2y^2\right)^2-13x^2y^2\)

\(=\left(2x^2-2y^2\right)^2-\left(\sqrt{13}xy\right)^2\)

\(=\left(2x^2-\sqrt{13}xy-2y^2\right)\left(2x^2+\sqrt{13}xy-2y^2\right)\)

Xem cách giải theo link: Câu hỏi của Trung Tính Hồ - Toán lớp 8 - Học toán với OnlineMath

Bài làm

a) Đặt t = x² + x + 1

bthuc <=> t( t + 1 ) - 12

            = t² + t - 12

            = t² - 3t + 4t - 12

            = t( t - 3 ) + 4( t - 3 )

            = ( t - 3 )( t + 4 )

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

            = ( x² + x - 2 )( x² + x + 5 )

            = ( x² - x + 2x - 2 )( x² + x + 5 )

            = [ x( x - 1 ) + 2( x - 1 ) ]( x² + x + 5 )

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

Bài làm

b) 10n² + n - 10

= 10n² - 10n + 9n - 9 - 1

= ( 10n² - 10n ) + ( 9n - 9 ) - 1

= 10n( n - 1 ) + 9( n - 1 ) - 1

= ( n - 1 )( 10n + 9 ) - 1

Ta có ( n - 1 )( 10n + 9 ) chia hết ( n - 1 )

=> Để ( 10n² + n - 10 ) chia hết ( n - 1 )

thì -1 chia hết ( n - 1 )

hay ( n - 1 ) ∈ Ư(-1) = { ±1 }

=> n ∈ { 2 ; 0 }