a) \({x^2} + \dfrac{1}{4}{x^2} - 5{x^2} = (1 + \dfrac{1}{4} - 5){x^2} =  - \dfrac{{15}}{4}{x^2}\);

b) \({y^4} + 6{y^4} - \dfrac{2}{5}{y^4} = (1 + 6 - \dfrac{2}{5}){y^4} = \dfrac{{33}}{5}{y^4}\).

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

\(=x^3+2^3-2\left(x^2-1\right)\)

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

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

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

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

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

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

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

\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)

\(=2x^3-x+x^2-y\)

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

giúp mk với nhé

sáng mai nộp rồi 

ai nhanh tay mk sẽ k cho

\(=\dfrac{x^2-4y}{xy}\cdot\dfrac{x^2}{x-y}=\dfrac{x\left(x^2-4y\right)}{y\left(x-y\right)}\)

Ta có:(x²-y²)\(.\dfrac{x^2+y^2}{y^4-x^2y^2}\)\(=\left(x^2-y^2\right).\dfrac{x^2+y^2}{y^2\left(y^2-x^2\right)}=-\dfrac{x^2+y^2}{y^2}\)

Ta có:\(\dfrac{4x^2-9y^2}{xy}:\left(2x-3y\right)=\dfrac{\left(2x-3y\right)\left(2x+3y\right)}{xy}.\dfrac{1}{\left(2x-3y\right)}=\dfrac{2x+3y}{xy}\)

a: \(=\dfrac{5}{2x^2y}+\dfrac{2}{3xy}-\dfrac{y}{x^3}\)

\(=\dfrac{5\cdot3\cdot x}{6x^3y}+\dfrac{2\cdot2\cdot x^2}{6x^3y}-\dfrac{6y^2}{6x^3y}\)

\(=\dfrac{15x+4x^2-6y^2}{6x^3y}\)

b: \(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)

c: \(=\dfrac{x^4-1-x^4+3x^2}{x^2-1}=\dfrac{3x^2-1}{x^2-1}\)