192=(20−1)2=202−2.20.1+12=400−40+1=361

a)

19^2=(20−1)^2=20^2−2.20.1+1^2=400−40+1=361

28^2=(30−2)^2=30^2−2.30.2+2^2=900−120+4=784

81^2=(80+1)^2=80^2+2.80.1+1^2=6400+160+1=6561

91^2=(90+1)^2=90^2+2.90.1+1^2=8100+180+1=8281

a) x⁶ + 1 

= (x²)³ + 1³

=(x² +1)(x⁴ - x² + 1)

b) x⁶ - y⁶

= (x²)³ - (y²)³

=(x² -y²)(x⁴ + x²y² + y⁴)

= (x-y)(x+y)[ x⁴ + 2.x²y² + y⁴ - x²y² ]

=(x-y)(x+y) [(x² +y²)² - (xy)² ]

=(x-y)(x+y)(x² - xy + y²)(x² + xy + y²)

c) x⁹ +1

= (x³)³ + 1³

=(x³ +1)(x⁶ - x³ -1)

= (x+1)(x² - x +1)(x⁶ - x³-1) 

D = (2x-1)³  -2x(2x-3)(2x+3) + 13x(x-1)

D= (2x)³ - 3. (2x)² .1 + 3.2x .1² -1³ - 2x(2x-3)(2x+3) + 13x(x-1)

D= 8x³ - 12x² + 6x -1 - 2x(4x² -9) + 13x² -13x

D= 8x³ -12x² + 6x-1 - 8x³ + 18x + 13x² -13x

D= (8x³ - 8x³) -(12x² -13x²) + (6x + 18x -13x) - 1

D= x² + 11x -1 

D = x² + 2x . 11/2 +(11/2)² -125/4

D= (x+ 11/2)² - 125/4

Với mọi x thì (x+11/2)² >= 0

=> (x+11/2)² - 125/4 >= -125/4

Dấu bằng xảy ra khi: (x+11/2)² =0

=> x + 11/2 =0 

=> x= -11/2

Vậy giá trị nhỏ nhất của D là -125/4 khi x= -11/2

c) (xy^2+1)^2

d) (1/3-y^4)^2

e) (1/2a-2b^2)^2

f) (5-x)^2

c) 2xy² + x² y⁴ + 1

= (xy²)² + 2xy² .1 + 1²

=(xy² +1)²

d) 1/9 - 2/3 y⁴ + y⁸

= (1/3)²  - 2 . 1/3 y⁴ + (y⁴)²

=(1/3 - y⁴)²

e) 1/4 a² - 2ab²  + 4b⁴

= (1/2 a)² - 2 1/2 a . 2b  + (2b²)²

=(1/2 a  - 2b²)²

f) 25 - 10x + x²

= x² + 2 . 5x + 5²

= (x+5)²

a, \(\frac{1}{25}x^2-\frac{1}{9}y^2=\left(\frac{1}{5}x\right)^2-\left(\frac{1}{3}y\right)^2=\left(\frac{1}{5}x-\frac{1}{3}y\right)\left(\frac{1}{5}x+\frac{1}{3}y\right)\)

b, \(-\left(4x^2-20x+25\right)=-\left[\left(2x\right)^2-2.2x.5+5^2\right]=-\left(2x-5\right)^2\)

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

e, \(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)

3x³ + 10x² + 2 = 3x³ + x² + 9x² + 3x - 3x - 1 + 3

= x²( 3x + 1 ) + 3x( 3x + 1 ) - ( 3x + 1 ) + 3

= ( 3x + 1 )( x² + 3x - 1 ) + 3

Vì ( 3x + 1 )( x² + 3x - 1 ) ⋮ ( 3x + 1 )

=> 3 ⋮ ( 3x + 1 ) <=> ( 3x + 1 ) ∈ Ư(3) ( đến đây bạn tự xét giá trị nhé )

Trả lời:

a, \(x^2=6x\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

Vậy x = 0; x = 6 là nghiệm của pt.

b, \(x^2-4x+3=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}}\)

Vậy x = 1; x = 3 là nghiệm của pt.

Bài 3:

a) \(x^2=6x\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

Vậy \(S=\left\{0;6\right\}\)

b) \(x^2-4x+3=0\)

\(\Leftrightarrow x^2-4x+4-1=0\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy \(S=\left\{3;1\right\}\)