\(a^3-3ab^2=5\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)

\(b^3-3a^2b=10\Rightarrow b^6-6a^2b^4+9a^4b^2=100\)

Cộng vế với vế:

\(a^6+3a^4b^2+3a^2b^4+b^6=125\)

\(\Leftrightarrow\left(a^2+b^2\right)^3=125\)

\(\Rightarrow a^2+b^2=5\)

\(\Rightarrow S=10090\)

\(\left(a^3-3ab^2\right)^2=25\Leftrightarrow a^6-6a^4b^2+9a^2b^4=25\)

\(\left(b^3-3a^2b\right)^2=100\Leftrightarrow b^6-6a^2b^4+9a^4b^2=100\)

\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125\)

\(\Leftrightarrow\left(a^2+b^2\right)^2=125\Leftrightarrow a^2+b^2=5\)

Thay a²+b²=5 vào S=2018a²+2018b²=2018(a²+b²)=2018.5=10090

Ta có :

+) \(a^3-3ab^2=5\Leftrightarrow\left(a^3-3ab^2\right)^2=25\Leftrightarrow a^6-6a^4b^2+9a^2b^4=25\)

+) \(b^3-3a^2b=10\Leftrightarrow\left(b^3-3a^2b\right)^2=100\Leftrightarrow b^6-6a^2b^4+9a^4b^2=100\)

\(\Leftrightarrow a^6+b^6+3a^2b^4+3a^4b^2=125\)

\(\Leftrightarrow\left(a^2+b^2\right)^3=125\)

\(\Leftrightarrow a^2+b^2=5\)

Ta cos :

\(S=2018a^2+2018b^2=2018\left(a^2+b^2\right)=2018.5=10090\)

Vaayj...

Bài 1 : 

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

Ta có : \(x+y+z=0\Rightarrow x+y=-z;y+z=-x;x+z=-y\)

hay \(-z.\left(-x\right)\left(-y\right)=-zxy\)

mà \(xyz=2\Rightarrow-xyz=-2\)

hay N nhận giá trị -2 

Bài 2 : 

\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)Đặt \(a=10k;b=3k\)

hay \(\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)

hay biểu thức trên nhận giá trị là 24 

c, Ta có : \(a-b=3\Rightarrow a=3+b\)

hay \(\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{-5+b}{b-5}-\frac{12+4b-b}{9+3b+3}\)

\(=\frac{-5+b}{b-5}-\frac{12+3b}{6+3b}\)quy đồng lên rút gọn, đơn giản rồi 

1.Ta có:\(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

\(\Rightarrow N=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\left(-x\right)\left(-y\right)=-2\)

2.Ta có:\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)

Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow a=10k;b=3k\)

Ta có:\(A=\frac{3a-2b}{a-3b}=\frac{3.10k-2.3k}{10k-3.3k}=\frac{30k-6k}{10k-9k}=\frac{k\left(30-6\right)}{k\left(10-9\right)}=24\)

Vậy....

Có : (a^3-3ab^2)^2 = 5^2 = 25

<=> a^6-6a^4b^2+9a^2b^2 = 25

(b^3-3a^2b)^2 = 10^1 = 100

<=> b^6-6a^2b^4+9a^4b^2 = 100

=> 5^3 = 125 = 25 + 100 = a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2 = a^6+b^6+3a^2b^4+3a^4b^2 = (a^2+b^2)^3

=> a^2+b^2 = 5

=> D = 312018.(a^2+b^2) = 312018 . 5 = 1560090

Tk mk nha

tk bn nha nhưng mà (a^3 - 3ab^2)^2 = a^6 - 9a^2b^4 mà bn