Ta  có : 99ⁿ  + 99^{n + 1} = 99ⁿ + 99ⁿ.99 = 99ⁿ(99 + 1) = 99ⁿ . 100\(⋮\)100 (đpcm)

P = 2x² + 12x + 9

= 2x² + 12x + 18 - 9

= 2( x² + 6x + 9 ) - 9

= 2( x + 3 )² - 9 ≥ -9 ∀ x

Dấu "=" xảy ra khi x = -3

=> MinP = -9 <=> x = -3

\(P=2x^2+12x+9=2\left(x^2+6x+9\right)-9\)

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

Vì \(2\left(x+3\right)^2\ge0\forall x;2\left(x+3\right)^2-9\ge-9\forall x\)

Vậy GTNN là -9 <=> x + 3 = 0 <=> x = -3

banj vieets laij deef ddi

\(\frac{1-4x^2}{x^2+4x}\div\frac{2-4x}{3x}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-4\\x\ne\frac{1}{2}\end{cases}}\)

\(=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\times\frac{3x}{2\left(1-2x\right)}\)

\(=\frac{3x\left(1+2x\right)}{2x\left(x+4\right)}\)

\(=\frac{3\left(1+2x\right)}{2\left(x+4\right)}=\frac{6x+3}{2x+8}\)

\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

\(=\left(x^2+4x\right)\left(x^2+16x+60\right)+128\)

\(=x^4+16x^3+60x^2+4x^3+64x^2+240x+128\)

\(=x^4+20x^3+124x^2+240x+128\)

x( x + 4 )( x + 6 )( x + 10 ) + 128

= [ x( x + 10 ) ][ ( x + 4 )( x + 6 ) ] + 128

= ( x² + 10x )( x² + 10x + 24 ) + 128

Đặt t = x² + 10x

= t( t + 24 ) + 128

= t² + 24t + 128

= t² + 8t + 16t + 128

= t( t + 8 ) + 16( t + 8 )

= ( t + 8 )( t + 16 )

= ( x² + 10x + 8 )( x² + 10x + 16 )

= ( x² + 10x + 8 )( x² + 2x + 8x + 16 )

= ( x² + 10x + 8 )[ x( x + 2 ) + 8( x + 2 ) ]

= ( x² + 10x + 8 )( x + 2 )( x + 8 )

nếu bn đưa = bao nhiêu mới giải đc

Bài 1 : 

a, sai đề 

b, \(\frac{x^2-xy-x+y}{x^2+xy-x-y}=\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\frac{\left(x-1\right)\left(x-y\right)}{\left(x-1\right)\left(x+y\right)}=\frac{x-y}{x+y}\)

Bài 2 : 

a, \(\frac{3x-7x}{3x-5x}-\frac{4x-7}{3x-5}=2-\frac{4x-7}{3x-5}=\frac{2\left(3x-5\right)-4x-7}{3x-5}\)

\(=\frac{6x-10-4x-7}{3x-5}=\frac{2x-17}{3x-5}\)

b, \(\frac{x^3-1}{x^3+x^2+x}-\frac{x^2-4}{6y-3xy}=\frac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2+x+1\right)}-\frac{\left(x-2\right)\left(x+2\right)}{3y\left(2-x\right)}\)

\(=\frac{x-1}{x}-\frac{-\left(2-x\right)\left(x+2\right)}{3y\left(2-x\right)}=\frac{x-1}{x}+\frac{x+2}{3y}\)