Ta có x = 99

=> x + 1 = 100

Khi đó A = x⁵ - 100x⁴ + 100x³ - 100x² + 100x - 9

= x⁵ - (x + 1)x⁴ + (x + 1)x³ - (x + 1)x² + (x + 1)x - 9

= x⁵ - x⁵ - x⁴ + x⁴ + x³ - x³ - x² + x² + x - 9

= x - 9 

Thay x = 99 vào A 

=> A = x - 9 = 99 - 9 = 90

Vậy A = 90

Ta có : \(x=99\Rightarrow100=x+1\)

\(A=x^5-100x^4+100x^3-100x^2+100x-9\)

\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)

\(=x-9\)hay \(99-9=90\)

Vậy \(A=90\)

Giúp tui câu b đi 😢

a, Với m = 3 ta được : 

<=> \(f\left(x\right)=2x^3+5x^2+5x+3\)

Ta có : \(f\left(x\right)⋮h\left(x\right)\)hay \(2x^3+5x^2+5x+3⋮x+1\)

2x^3 + 5x^2 + 5x + 3 x + 1 2x^2 + 3x + 2 2x^3 + 2x^2 3x^2 + 5x 3x^2 + 3x 2x + 3 2x + 2 1

b, 

2x^3 + 5x^2 + 5x + m x + 1 2x^2 + 3x + 2 2x^3 + 2x^2 3x^2 + 5x 3x^2 + 3x 2x + m 2x + 2 m - 2

Để m - 2 = 0 <=> m = 2

\(49-4x^2+2x+7\)

\(=-4x^2+2x+56\)

\(=-2\left(2x^2-x-28\right)\)

\(=-2\left(2x^2-8x+7x-28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]\)

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

\(49-4x^2+2x+7=-4x^2+2x+56\)

\(=-2\left(2x^2-x+28\right)=-2\left(2x^2-8x+7x+28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]=-2\left(2x+7\right)\left(x-4\right)\)

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Bài làm

Ta có : a³ + b³ + c³ = 3abc

<=> ( a³ + b³ ) + c³ - 3abc = 0

<=> ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

<=> [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

<=> ( a + b + c )[ ( a + b )² - ( a + b )c + c² ] - 3ab( a + b + c ) = 0

<=> ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

<=> ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Vì a, b, c dương => a + b + c > 0 => a + b + c = 0 vô lí

Xét a² + b² + c² - ab - bc - ac = 0

<=> 2( a² + b² + c² - ab - bc - ac ) = 2.0

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

<=> ( a - b )² + ( b - c )² + ( a - c )² = 0

VT ≥ 0 ∀ a,b,c . Đẳng thức xảy ra <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}}\Leftrightarrow a=b=c\)

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

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

\(=0\)

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