ĐKXĐ: \(x\ge1;y\ge25\)

\(D=\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}+\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\)

Vì x>=1,y>=25 => x-1>=0,y-25>=0 

=> D >= 0

Dấu "=" xảy ra <=> x=1,y=25

Vậy MinD=0 khi x=1,y=25

Ta có: \(\left(x-2\right)^2+25\ge25;\left(y-50\right)^2+1\ge1\)

=>\(\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}\le\frac{1}{x}\sqrt{\frac{x-1}{25}};\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\le\frac{1}{y}\sqrt{y-25}\)

=>\(D\le\frac{1}{x}\sqrt{\frac{x-1}{25}}+\frac{1}{y}\sqrt{y-25}\)

Vì x>=1 => x-1>=0. Áp dụng bđt cosi với 2 số dương x-1 và 1 ta có:

\(\sqrt{x-1}=\sqrt{\left(x-1\right).1}\le\frac{x-1+1}{2}=\frac{x}{2}\)

=>\(\frac{1}{x}\sqrt{\frac{x-1}{25}}\le\frac{1}{x}\cdot\frac{x}{2}\cdot\frac{1}{\sqrt{25}}=\frac{1}{10}\)

Vì y>=25 => y-25>=0. ÁP dụng bđt cô si cho 2 số dương 25 và y-25 ta có:

\(\sqrt{y-25}=\frac{\sqrt{25\left(y-25\right)}}{5}\le\frac{25+y-25}{2.5}=\frac{y}{10}\)

=>\(\frac{1}{y}\sqrt{y-25}=\frac{1}{y}\cdot\frac{y}{10}=\frac{1}{10}\)

Suy ra \(D\le\frac{1}{10}+\frac{1}{10}=\frac{1}{5}\)

Dấu "=" xảy ra <=> x=2,y=50

Vậy MaxD = 1/5 khi x=2,y=50

Akai Haruma Bonking

a) 3x³ + 6x²y

= 3x².(x + 2y)

b) 2x³ - 6x²

= 2x².(x - 2)

c) 18x² - 20xy

= 2x.(9x - 10y)

d) xy + y² - x - y

= (xy + y²) - (x + y)

= y(x + y) - (x + y)

= (x + y)(y - 1)

e) (x²y² - 8)² - 1

= (x²y² - 8 - 1)(x²y² - 8 + 1)

= (x²y² - 9)(x²y² - 7)

= (xy - 3)(xy + 3)(x²y² - 7)

f) x² - 7x - 8

= x² - 8x + x - 8

= (x² - 8x) + (x - 8)

= x(x - 8) + (x - 8)

= (x - 8)(x + 1)

a: \(3x^3+6x^2y\)

\(=3x^2\cdot x+3x^2\cdot2y=3x^2\left(x+2y\right)\)

b: \(2x^3-6x^2=2x^2\cdot x-2x^2\cdot3=2x^2\left(x-3\right)\)

c: \(18x^2-20xy=2x\cdot9x-2x\cdot10y=2x\left(9x-10y\right)\)

d: \(xy+y^2-x-y\)

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

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

e: \(\left(x^2y^2-8\right)^2-1\)

\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)

\(=\left(x^2y^2-7\right)\left(x^2y^2-9\right)\)

\(=\left(x^2y^2-7\right)\left(xy-3\right)\left(xy+3\right)\)

f: \(x^2-7x-8\)

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

\(=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)

g: \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)

\(=2x\cdot\left(2x-y\right)\cdot5x-2x\cdot\left(2x-y\right)\cdot3y\)

\(=2x\left(2x-y\right)\left(5x-3y\right)\)

h: \(x^2-2x+1-y^2\)

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

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

i: \(2x\left(x+2\right)+x^2\left(-x-2\right)\)

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

k: \(-x^2+6x-9=-\left(x^2-6x+9\right)\)

\(=-\left(x^2-2\cdot x\cdot3+3^2\right)=-\left(x-3\right)^2\)

l: \(-2x^2+8xy-8y^2\)

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

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

m: \(3x^2+5x-3y^2-5y\)

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

\(=3\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x+3y+5\right)\)

g) 10x²(2x - y) + 6xy(y - 2x)

= 10x²(2x - y) - 6xy(2x - y)

= 2x(2x - y)(5x - 3y)

h) x² - 2x + 1 - y²

= (x² - 2x + 1) - y²

= (x - 1)² - y²

= (x - y - 1)(x + y - 1)

i) 2x(x + 2) + x² (-x - 2)

= 2x(x + 2) - x²(x + 2)

= x(x + 2)(2 - x)

k) -9 + 6x - x²

= -(x² - 6x + 9)

= -(x - 3)²

l) 8xy - 2x² - 8y²

= -2(x² - 4xy + 4y²)

= -2(x - 2y)²

m) 3x² + 5x - 3y² - 5y

= (3x² - 3y²) + (5x - 5y)

= 3(x² - y²) + 5(x - y)

= 3(x - y)(x + y) + 5(x - y)

= (x - y)[3(x + y) + 5]

= (x - y)(3x + 3y + 5)

chờ xíu đang ghi nha

a) \(x^3+3.2x^2y+3.2^2.x.y^2+\left(2y\right)^3=\left(x+2y\right)^3\)

b) áp dụng HDT : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

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

c) cũng áp dụng hdt :\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2=\left[3\left(x+5\right)-x+7\right]\left[3\left(x+5\right)+x-7\right]\)\(=\left(3x+15-x+7\right)\left(2x+15+x-7\right)=\left(2x+22\right)\left(3x+8\right)=2\left(x+11\right)\left(3x+8\right)\)

d) áp dụng típ \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)

\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)=\left(x-9y\right)\left(9x-y\right)\)

e)Áp dụng típ Hdt như trên

\(\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)=\left(4y-34\right)\left(11y-22\right)\)

\(=2\left(2y-17\right).11\left(y-2\right)=22\left(2y-17\right)\left(y-2\right)\)

Bạn 1 cái t i c k nha thật sự rất cảm ơn

a) Ta có x² + 9y² - 6xy = (x - 3y)² (1)

Thay x = 16 ; y = 2 vào (1) ta có

(x - 3y)² = (16 - 2.3)² = 10² = 100

b) Ta có x³ - 6x²y + 12xy² - 8y³

= (x - 2y)³ (1)

Thay x = 14 ; y = 2 vào  (1) ta có 

(x - 2y)³ = (14 - 2.2)³ = 10³ = 1000 

a) \(x^2+9y^2-6xy=\left(x-3y\right)^2\)

Thay \(x=16;y=2\)vào biểu thức trên ta có :

\(\left(16-3.2\right)^2=\left(16-6\right)^2=10^2=100\)

Vậy tại x = 16 và y = 2 thì biểu thức trên = 100

b) \(x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

Thay x = 14 và y = 2 vào biểu thức trên ta có :

\(\left(14-2.2\right)^3=\left(14-4\right)^3=10^3=1000\)

Vậy tại x = 14 và y = 2 thì biểu thức trên = 1000

\(a,x^2+9y^2-6xy\)

\(=x^2-6xy+9y^2\)

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

Tại x=16 và y=2 ta có:

\(\left(x-3y\right)^2=\left(16-3.2\right)^2=100\)

\(b,x^3-6x^2y+12xy^2-8y^3\)

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

Tại x=14 và y=2 ta có

\(\left(x-2y\right)^3=\left(14-2.2\right)^3=1000\)

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

ỳtct7ct7c7c7t79tc9

\(3,x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\\ 4,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\\ 5,x^2-2xy+y^2-xz+yz=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y-z\right)\left(x-y\right)\\ 6,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\\ 9,x^3+x^2-xy+xy+y^2+y^3\\ =x^2\left(x+1\right)+y^2\left(x+1\right)=\left(x^2+y^2\right)\left(x+1\right)\\ 10,x^2-6\left(x+3\right)-9\\ =x^2-6x-18-9\\ =x^2-6x-27=\left(x-9\right)\left(x+3\right)\)

10: \(x^2-6\left(x+3\right)-9\)

\(=x^2-6x-18-9\)

\(=x^2-6x-27\)

\(=\left(x-9\right)\left(x+3\right)\)

3)(x-1).(x-y)