Câu 2 :
2) Ta có : \(\Rightarrow x=-y-z\)
\(\Rightarrow x.x=x.\left(-y-z\right)=-xy-yz\)
\(\Rightarrow x^2=-xy-yz\)
\(\Rightarrow2x^2+yz=x^2+yz-xy-yz\)
\(=x.\left(x-y\right)-z.\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự ta có : \(2y^2+zx=\left(y-z\right)\left(y-x\right)\)
\(2z^2+xy=\left(z-x\right)\left(z-y\right)\)
Khi đó ta có : \(A=\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\dfrac{-x^2.\left(y-z\right)-y^2.\left(z-x\right)-z^2.\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{x^2\left(z-y\right)-y^2.\left(z-x\right)+z^2.\left(y-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{x^2\left(z-y\right)-y^2.\left[\left(z-y\right)+\left(y-x\right)\right]+z^2.\left(y-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{x^2\left(z-y\right)-y^2\left(z-y\right)-y^2.\left(y-x\right)+z^2.\left(y-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{\left(z-y\right)\left(x^2-y^2\right)+\left(y-x\right)\left(z^2-y^2\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{\left(z-y\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(z-y\right)\left(z+y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{\left(z-y\right)\left(x-y\right)\left(x+y-z-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{\left(z-y\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Vậy \(A=1\)
bài 3 chứng minh rằng nếu a,b là các số nguyên tố lớn hơn 2 thì a^3b- ab^2 cha hết cho 240
bài 3 chứng minh rằng nếu a,b là các số nguyên tố lớn hơn 2 thì a^3b- ab^2 chia hết cho 240
Mọi người giúp e bài 17 vs ạ
