Ta có: 

\(P=\dfrac{5x-4y}{5x+4y}\)

\(\Leftrightarrow P^2=\left(\dfrac{5x-4y}{5x+4y}\right)^2\)

\(\Leftrightarrow P^2=\dfrac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}\)

\(\Leftrightarrow P^2=\dfrac{\left(5x\right)^2-2\cdot5x\cdot4y+\left(4y\right)^2}{\left(5x\right)^2+2\cdot5x\cdot4y+\left(4y\right)^2}\)

\(\Leftrightarrow P^2=\dfrac{\left(25x^2+16y^2\right)-40xy}{\left(25x^2+16y^2\right)+40xy}\)

Thay \(25x^2+16y^2=50xy\) vào ta có:

\(P^2=\dfrac{50xy-40xy}{50xy+40xy}=\dfrac{10xy}{90xy}=\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2\)

Mà: \(4y< 5x< 0\)

Nên: \(P=\dfrac{5x-4y}{5x+4y}< 0\)

Vậy: \(P=-\dfrac{1}{3}\)

25x^2+16y^2=50xy

=>25x^2-50xy+16y^2=0

=>25x^2-10xy-40xy+16y^2=0

=>5x(5x-2y)-8y(5x-2y)=0

=>(5x-2y)(5x-8y)=0

=>5x=2y hoặc 5x=8y

5x>4y

=>5x=8y

=>x/8=y/5=k

=>x=8k; y=5k

\(P=\dfrac{5\cdot8k-4\cdot5k}{5\cdot8k+4\cdot5k}=\dfrac{40-20}{40+20}=\dfrac{1}{3}\)

`a)100x^2-20x+1`

`=(10x-1)^2`

Thay `x=1/10`

`=>100x^2-20x+1=(1-1)^2=0`

`b)49x^2-42x+10`

`=49*4/49-42*2/7+10`

`=4-12+10=2`

`c)25x^2+40x+16y^2`

`=(5x+4y)^2=(2+3)^2=25`

(3x-4y)²

\(\left(3x-4y\right)^2\)

9x² - 24xy + 16y²

= (3x - 4y)²

\(x^2-25=\left(x-5\right)\left(x+5\right)\\ x^2+10x+25=\left(x+5\right)^2\\ x^2-6x+xy-6y=x\left(x-6\right)+y\left(x-6\right)=\left(x+y\right)\left(x-6\right)\\ x^2-2x-y^2+1=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)

\(a,4x^2+28x+49=\left(2x\right)^2+2.2x.7+7^2=\left(2x+7\right)^2\\ b,16y^2-8y+1=\left(4y\right)^2-2.4y.1+1^2=\left(4y-1\right)^2=\left(1-4y\right)^2\\ 4a^2+20ab+25b^2=\left(2a\right)^2+2.2a.5b+\left(5b\right)^2=\left(2a+5b\right)^2\\ d,9x^2-6xy+y^2=\left(3x\right)^2-2.3x.y+y^2=\left(3x-y\right)^2=\left(y-3x\right)^2\)

\(4xy^2-6+8y\)

\(4xy^2-6+8y\)

a, \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)

b, \(4m^2-16n^2=\left(2m-4n\right)\left(2m+4n\right)=4\left(m-2n\right)\left(m+2n\right)\)

c, \(49-16x^2=\left(7-4x\right)\left(7+4x\right)\)

d, \(25-9y^2=\left(5-3y\right)\left(5+3y\right)\)

e, \(81x^2-16y^2=\left(9x-4y\right)\left(9x+4y\right)\)

b: \(8x^2-48x+6xy-36y\)

\(=8x\left(x-6\right)+6y\left(x-6\right)\)

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

d: \(a^2-2ab+b^2-4\)

\(=\left(a-b\right)^2-4\)

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

D   =   ( 15 x y 2   +   18 x y 3   +   16 y 2 )   :   6 y 2   –   7 x 4 y 3   :   x 4 y     ⇔   D   =   15 x y 2   :   ( 6 y 2 )   +   18 x y 3   :   ( 6 y 2 )   +   16 y 2   :   ( 6 y 2 )   –   7 x 4 y 3   :   x 4 y     ⇔   D   = 5 2 x + 3 x y + 8 3 - 7 y 2

Tại x = 2 3  và y = 1 ta có

  D   = 5 2 . 2 3 + 3 . 2 3 . 1 + 8 3 - 7 . 1 2 = 5 3 + 2 + 8 3 - 7 = 13 3 - 5 = - 2 3  

Đáp án cần chọn là: D

a) \(a^2b-4ab^2\)

\(=ab\left(a-4b\right)\)

b) \(x^8+4y^4\)

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

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

c) \(x^2-10x-16y^2+25\)

\(=\left(x^2-10x+25\right)-16y^2\)

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

\(=\left[\left(x-5\right)+4y\right]\left[\left(x-5\right)-4y\right]\)

\(=\left(x-5+4y\right)\left(x-5-4y\right)\)