\(\frac{7}{4}-y.\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)

\(\Leftrightarrow\frac{5}{6}.y=\frac{7}{4}-\frac{1}{2}-\frac{1}{3}\)

\(\Leftrightarrow\frac{5}{6}.y=\frac{11}{12}\)

\(\Leftrightarrow y=\frac{11}{12}:\frac{5}{6}\)

\(\Leftrightarrow y=\frac{11}{12}.\frac{6}{5}\)

\(\Leftrightarrow y=\frac{11}{10}\)

Vậy\(y=\frac{11}{10}\)

Dumflinz

Có:\(x^4+64y^4\)

\(=\left(x^4+16x^2y^2+64y^4\right)-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+4xy+8y^2\right)\left(x^2-4xy+8y^2\right)\)

Linz

= 64y⁴ + 32xy³ + 8y²x²  - 32xy³  -16x²y² -  4x³y + 8x²y² +4x³y +x⁴

= 8y² ( 8y² + 4xy + x² ) - 4xy ( 8y² + 4xy + x² ) + x²  ( 8y² + 4xy + x² )

= ( 8y² - 4xy + x² ) ( 8y² + 4xy + x² )

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

\(=\frac{x^3}{xyz}+\frac{y^3}{xyz}+\frac{z^3}{xyz}\)

\(=\frac{x^3+y^3+y^3}{xyz}\)

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)-3xyz}{xyz}\)( HĐT x³ + y³ + z³ - 3xyz chắc biết rồi ha )

\(=\frac{-3xyz}{xyz}=-3\)( do x+y+z = 0 )

à nhầm dấu tí nhé ._.

\(A=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz}{xyz}=3\)

A = x² - 5x + 7 

= ( x² - 5x + 25/4 ) + 3/4

= ( x - 5/2 )² + 3/4 ≥ 3/4 ∀ x 

Dấu "=" xảy ra khi x = 5/2

=> MinA = 3/4, đạt được khi x = 5/2