3 cho a,b,c > 0 và \(a^2b^2c^2=3\) C/m \(\dfrac{a^3}{bc}\dfrac{b^3}{ca}\dfrac{c^3}{ab} {a^3}{a^2b^2}\dfrac{b^3}{b^2c^2}\dfrac{c^3}{c^2a^2}>=\dfrac{abc}{2}\) Theo dõi Vi phạm YOMEDIA Toán 10 Chương 4 Bài 1 Trắc nghiệm Toán 10 Chương 4 Bài 1 Giải bài tập Toán 10 Chương 4 Bài 1 Trả lời (1) Bài 3Click here👆to get an answer to your question ️ Simplify (a^2b^2)^3(b^2c^2)^3(c^2a^2)^3/(a b)^3(bc)^3(ca)^3Consider a triangle with sides of length a, b, c, where θ is the measurement of the angle opposite the side of length cThis triangle can be placed on the Cartesian coordinate system aligned with edge a with origin at C, by plotting the components of the 3 points of the triangle as shown in Fig 4 = (, ), = (,), = (,) By the distance formula, = (− ) (− ) If Ab Ca 4 And Abc 2 Can You Find The Value Of 1 A 1 B 1 C Quora (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3/(a-b)^3+(b-c)^3+(c-a)^3
