a: \(\frac{x}{y}=\frac13\)

=>y=3x

\(\frac{14x+5y}{3x-11y}=\frac{11x+5\cdot3x}{3x-11\cdot3x}=\frac{26x}{3x-33x}=\frac{26x}{-30x}=\frac{-26}{30}=-\frac{13}{15}\)

b: \(\frac{a}{b}=\frac12\)

=>b=2a

\(\frac{11a^4-3ab^3+15a^3b+7b^4}{3a^2b^2+ab^3-6a^3b-2b^4}\)

=\(\frac{11a^4-3a\cdot\left(2a\right)^3+15a^3\cdot2a+7\left(2a\right)^4}{3a^2\cdot\left(2a\right)^2+a\cdot\left(2a\right)^3-6a^3\cdot2a-2\cdot\left(2a\right)^4}\)

\(=\frac{11a^4-3a\cdot8a^3+30a^4+7\cdot16a^4}{3a^2\cdot4a^2+a\cdot8a^3-6a^3\cdot2a-2\cdot16a^4}\)

\(=\frac{11a^4-24a^4+30a^4+112a^4}{12a^4+8a^4-12a^4-32a^4}=\frac{11-24+30+112}{12+8-12-32}=\frac{129}{-24}=\frac{-43}{8}\)

a) 4x^2 - 12xy + 9y^2

=(2x)^2 - 2.2.3xy + (3y)^2

=(2x+3y)^2

b) 27a^3 - 64b^3

=(3a)^3 - (4b)^3

=(3a - 4b) [(3a)^2 +3a.4b +(4B)^2]

d) (2x - 6y)^2 - (3xy - 4)^2

=[ (2x - 6y)+ (3xy - 4) ] [ (2x - 6y)- (3xy - 4) ]

\(1,a,4x^2-12xy+9y^2\)

\(=\left(2x\right)^2-2.3.2xy+\left(3y\right)^2\)

\(=\left(2x-3y\right)^2\)

\(b,27a^3-64b^3\)

\(=\left(3a\right)^3-\left(4b\right)^3\)

\(\left(3a-4b\right)\left(9a^2+12ab+16b^2\right)\)

Ta luôn có 

\(x^2+2xy+y^2=\left(x+y\right)^2\) ( hẳng đẳng thức )

\(\Rightarrow A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(3a-2b\right)^2\)

\(=\left[\left(2a-3b\right)+\left(3a-2b\right)\right]^2\)

\(=\left(2a-3b-2b+3a\right)^2\)

\(=\left(a-b\right)^2\)

\(=10^2\)

\(=100\)

4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

con dc thầy tick 

thêm GP 

=))