\(a)\)\(x^2-y^2-2x+2y\)

\(=\left(x^2-2x+1\right)-\left(y^2-2y+1\right)\)

\(=\left(x-1\right)^2-\left(y-1\right)^2\)

\(=\left(x-y\right)\left(x+y-2\right)\)

\(b)\)\(x^2+4y^2-25+4xy\)

\(=\left(x^2+4xy+4y^2\right)-25\)

\(=\left(x+2y\right)^2-25\)

\(=\left(x+2y-5\right)\left(x+2y+5\right)\)

Dumflinz

Có:\(A=\frac{1}{x^2+4x+5}\)

         \(=\frac{1}{\left(x^2+4x+4\right)+1}\)

         \(=\frac{1}{\left(x+2\right)^2+1}\)

Vì\(\left(x+2\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x+2\right)^2+1\ge1\forall x\)

\(\Leftrightarrow\frac{1}{\left(x+2\right)^2+1}\le1\forall x\)

\(\Leftrightarrow A\le1\forall x\)

Dấu "=" xảy ra <=> x + 2 = 0

                        <=> x = -2

Vậy A đạt GTLN bằng 1 tại x = -2.

Dumflinz

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

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

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

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

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

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

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 )