Đặt f(x) = x⁴ + ax³ + b

      g(x) = x² - 1 = ( x - 1 )( x + 1 )

f(x) chia hết cho g(x) <=> x⁴ + ax³ + b chia hết cho ( x - 1 )( x + 1 )

<=> \(\hept{\begin{cases}\left(x^4+ax^3+b\right)⋮\left(x-1\right)\left[1\right]\\\left(x^4+ax^3+b\right)⋮\left(x+1\right)\left[2\right]\end{cases}}\)

Áp dụng định lí Bézout vào [1] ta có :

f(x) chia hết cho ( x - 1 ) <=> f(1) = 0

<=> 1 + a + b = 0

<=> a + b = -1 (1)

Áp dụng định lí Bézout vào [2] ta có :

f(x) chia hết cho ( x + 1 ) <=> f(-1) = 0

<=> 1 - a + b = 0

<=> -a + b = -1 (2)

Từ (1) và (2) => \(\hept{\begin{cases}a+b=-1\\-a+b=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=-1\end{cases}}\)

Vậy a = 0 ; b = -1

Ta có: \(a+b+c=ab+bc+ca\)

\(\Rightarrow a+b+c-ab-bc-ca=0\)

\(\Rightarrow a+b+c-ab-bc-ca+1-1=0\)

\(\Rightarrow a+b+c-ab-bc-ca+abc-1=0\)

\(\Rightarrow a\left(1-b\right)-\left(1-b\right)+c\left(1-b\right)-ca\left(1-b\right)=0\)

\(\Rightarrow\left(1-b\right)\left(a-1+c-ca\right)=0\)

\(\Rightarrow\left(1-b\right)\left[a\left(1-c\right)-\left(1-c\right)\right]=0\)

\(\Rightarrow\left(1-b\right)\left(1-c\right)\left(a-1\right)=0\)

=> 1 - b = 0      hoặc        1 - c = 0      hoặc    a - 1 = 0

=> b = 1                            c = 1                       a = 1

Vậy trong các số a, b, c có ít nhất một số bằng 1

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

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

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

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

\(=\frac{\left(y-z\right)\left(x^3-y^3\right)-\left(x-y\right)\left(y^3-z^3\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

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

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

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

\(=\frac{\left(x-y\right)\left(y-z\right)\left[\left(x-z\right)\left(x+z\right)+y\left(x-z\right)\right]}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(x+z+y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=x+y+z=2020\)

x^2+25=10x

=> x^2-10x+25=0

=> (x-5)^2=0

=>x-5=0

=>x=5

a, \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\Leftrightarrow x=\frac{1}{5};1\)

b, \(x^2+25=10x\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)