Question

[Trần Tuấn Trọng]

Cho xy+xz+yz =0 và  xyz \(\ne\)0 Tính M= \(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)

Answer 1

[OoO_Nhok_Lạnh_Lùng_OoO]

làm tương tự bài này nha

x + y + z = 3. Tìm Max P = xy + yz + xz

Ta có: (x - y)² ≥ 0 <=> x² - 2xy + y² ≥ 0 <=> x² + y² ≥ 2xy

hay 2xy ≤ x² + y² , dấu " = " xảy ra <=> x = y 

tương tự: 

+) 2yz ≤ y² + z² +) 2xz ≤ x² + z² 

cộng 3 vế của 3 bđt trên

--> 2xy + 2yz + 2xz ≤ 2(x² + y² + z²) 

--> xy + yz + xz ≤ x² + y² + z² 

--> xy + yz + xz + 2xy + 2yz + 2xz ≤ x² + y² + z² + 2xy + 2yz + 2xz 

--> 3(xy + yz + xz) ≤ (x + y + z)² 

--> 3(xy + yz + xz) ≤ 3² 

--> xy + yz + xz ≤ 3 

Answer 2

[Thúy Ngân]

Theo đề ta có :

xy + yz + xz = 0 

\(\Rightarrow xy=0-yz-xz=-\left(yz+xz\right)\) (1)

\(\Rightarrow yz=0-xz-xy=-\left(xz+xy\right)\)(2)

\(\Rightarrow xz=0-xy-yz=-\left(xy+yz\right)\)(3)

\(M=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)

Từ (1) ; (2) và (3) , ta có :

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

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

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

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

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

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

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

..............