olm hoi bai nhu cai dau buoi á

Tham khảo câu trả lời

1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)

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

2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)

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

3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)

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

4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)

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

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

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

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

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

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

\(=\left(x+y\right)\left(y+z\right)\left(x-z\right)\)

1) 4x² - 7x - 2 = 4x² - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )

2) 4x² + 5x - 6 = 4x² - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )

3) 5x² - 18x - 8 = 5x² - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )

4) xy( x + y ) - yz( y + z ) + xz( x - z )

= x²y + xy² - y²z - yz² + xz( x - z )

= ( x²y - yz² ) + ( xy² - y²z ) + xz( x - z )

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

= y( x - z )( x + z ) + y²( x - z ) + xz( x - z )

= ( x - z )[ y( x + z ) + y² + xz ]

= ( x - z )( xy + yz + y² + xz )

= ( x - z )[ ( xy + y² ) + ( xz + yz ) ]

= ( x - z )[ y( x + y ) + z( x + y ) ]

= ( x - z )( x + y )( y + z )

5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )

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

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

\(c,=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)

\(d,=x^2y^2-y^2-x^2+1=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)

\(e,4x^2+4x-15=\left(4x^2+4x+1\right)-16=\left(2x+1\right)^2-4^2=\left(2x+5\right)\left(2x-3\right)\)

\(3x^2-7x+2=\left(3x^2-6x\right)-\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)

\(4x^2-5x+1=\left(4x^2-4x\right)-\left(x-1\right)=4x\left(x-1\right)-\left(x-1\right)=\left(4x-1\right)\left(x-1\right)\)

Phân tích à :v

a) x³ + x² - 4x - 4 = x²( x + 1 ) - 4( x + 1 ) = ( x + 1 )( x² - 4 ) = ( x + 1 )( x - 2 )( x + 2 )

b) x⁴ + x³ + x² - 1 = x³( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 )

c) x² + 2xy + y² - 2x - 2y + 1 = ( x² + 2xy + y² ) - ( 2x + 2y ) + 1 = ( x + y )² - 2( x + y ) + 1² = ( x + y - 1 )²

d) x²y² + 1 - x² - y² = ( x²y² - x² ) - ( y² - 1 ) = x²( y² - 1 ) - ( y² - 1 ) = ( y² - 1 )( x² - 1 ) = ( y - 1 )( y + 1 )( x - 1 )( x + 1 )

e) 4x² + 4x - 15 = ( 4x² + 4x + 1 ) - 16 = ( 2x + 1 )² - 4² = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )

g) 3x² - 7x + 2 = 3x² - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )

h) 4x² - 5x + 1 = 4x² - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )

Chứng minh

a) \(2\equiv-1\left(mod3\right)\)

\(\Rightarrow2^{1000}\equiv\left(-1\right)^{1000}\equiv1\left(mod3\right)\Rightarrow2^{1000}-1\equiv0\left(mod3\right)\Rightarrowđpcm\)

b) \(19\equiv-1\left(mod20\right)\)

\(\Rightarrow19^{45}\equiv\left(-1\right)^{45}\equiv1\left(mod20\right);19^{30}\equiv\left(-1\right)^{30}\equiv1\left(mod20\right)\)

\(\Rightarrow19^{45}+19^{30}\equiv0\left(mod20\right)\Rightarrowđpcm\)

a) x³ - 6x² + 11x - 6

= ( x³ - 2x² ) - ( 4x² - 8x ) + ( 3x - 6 )

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

= ( x - 2 )( x² - 4x + 3 )

= ( x - 2 )( x² - x - 3x + 3 )

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

= ( x - 2 )( x - 1 )( x - 3 )

b) x³ - 6x² - 9x + 14

= ( x³ - x² ) - ( 5x² - 5x ) - ( 14x - 14 )

= x²( x - 1 ) - 5x( x - 1 ) - 14( x - 1 )

= ( x - 1 )( x² - 5x - 14 )

= ( x - 1 )( x² + 2x - 7x - 14 )

= ( x - 1 )[ x( x + 2 ) - 7( x + 2 ) ]

= ( x - 1 )( x + 2 )( x - 7 )

c) x³ + 6x² + 11x + 6

= ( x³ + 2x² ) + ( 4x² + 8x ) + ( 3x + 6 )

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

= ( x + 2 )( x² + 4x + 3 )

= ( x + 2 )( x² + x + 3x + 3 )

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

= ( x + 2 )( x + 1 )( x + 3 )

e) x⁶ - 9x³ + 8

Đặt t = x³

bthuc <=> t² - 9t + 8 

            = t² - t - 8t + 8

            = t( t - 1 ) - 8( t - 1 )

            = ( t - 1 )( t - 8 )

            = ( x³ - 1 )( x³ - 8 )

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

\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)

\(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

\(=\left(5x-1\right)\left(x+y\right)\)

Sửa đề : a, \(x\left(x-2\right)^2-\left(x-2\right)\left(x^2+2x+4\right)\)

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

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

x( x - 2 )² - ( x - 2 )( x² + 2x + 4 ) < bỏ cái ² đi nhá :v >

= x( x² - 4x + 4 ) - ( x³ - 8 )

= x³ - 4x² + 4x - x³ + 8

= -4x² + 4x + 8