a: \(=\dfrac{30xy^2}{7x^2y^3}=\dfrac{30}{7xy}\)

b: \(=\dfrac{10xy^2}{7x^2y^3}=\dfrac{10}{7xy}\)

\(=xy+2x^2+2y^2+2xy\)

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

\(\left(x+2y\right)\left(y+2x\right)=xy+2x^2+2y^2+4xy=2x^2+2y^2+5xy\)

\(=xy+2x^2+2y^2+4xy=2x^2+5xy+2y^2\)

\(\left(x+2y\right)\left(y+2x\right)=xy+2x^{^2}+2y^{^2}+4xy=5xy+2x^{^2}+2y^{^2}\)

1.

\(\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

\(=5x^2-3xy^2+4y\)

2.

a)  \(27x^4-8x=x\left(27x^3-8\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b)  \(16x^2y-4xy^2-4x^3+x^2y\)

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

\(=x\left(4x-y\right)\left(4y-x\right)\)

c) \(x^2-2x-5+2\sqrt{5}\)

\(=\left(x-1\right)^2-6+2\sqrt{5}\)

\(=\left(x-1\right)^2-\left(6-2\sqrt{5}\right)=\left(x-1\right)^2-\left(\sqrt{5}-1\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)\)

Bài 1:

 \(\left(25x^4y^3-15x^3y^5+20x^2y^4\right):\left(5x^2y^3\right)\)

\(=\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

\(=5x^2-3xy^2+4y\)

Bài 2: 

a) \(27x^4-8x\)

\(=x\left(3x-2\right)\left(3^2x^2+2.3x+2^2\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b) \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

Answer:

\(3x^2.\left(2x^3-x+5\right)\)

\(=3x^2.2x^3+3x^2.(-x)+3x^2.5\)

\(=6x^5-3x^3+15x^2\)

\((4xy+3y-5x).x^2y\)

\(=4xy.x^2y+3y.x^2y-5x.x^2y\)

\(=4x^3+3x^2y^2-5x^3y\)

Đáp án 8 xy x 2 − 4 y 2 .

\(x^2+4y^2-5x+10y-4xy+20\)

\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)

\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)

\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được : 

\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)

\(B=x^2-2xy-2x+2y+y^2\)

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

\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được : 

\(=1-2=-1\)

Đây em nhé!

a/ \(\Rightarrow\int^{4x-2y=2}_{-3x+2y=2}\)

Cộng 2 vế ta đc : x = 4

Thay x = 4 vào 2x - y = 1 ta đc:

8 - y = 1

=> y = 7

Vậy x = 4 ; y = 7

b/ \(\Rightarrow\int^{3x+4y=12}_{10x+4y=10}\)

Trừ 2 vế ta đc : 7x = -2 => x = -2/7

Thay x = -2/7 vào 3x + 4y = 12 ta đc :

-6/7 + 4y = 12 

=> 4y = 90/7

=> y = 45/14

Vậy x = -2/7 ; y = 45/14

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