chưa trả lời mà đã thanks

a/ 16x² - 4y² = (4x + 2y)(4x - 2y)

b/ x⁴ + 1 = (x² + √2 x + 1)(x² - √2 x + 1)

a/ 16x²-4y²=(4x)²-(2y)²=(4x+2y)(4x-2y)

b/ x⁴+1=(x²)²+1²=(x²+1)(x²-1)

 k minh nha

alibaba nguyễn sai câu b rồi \(\sqrt{2}\)làm sao bằng 1 đc

a/ \(x^2-4x+3=\left(x^2-x\right)-\left(3x-3\right)=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)

b/ \(3x^2-5x+2=\left(3x^2-3x\right)-\left(2x-2\right)=3x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(3x-2\right)\)

\(x^2-4x+3\)

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

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

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

\(3x^2-5x+2\)

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

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

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

a/ x²-4x+3=x²-x-3x+3=x(x-1)-3(x-1)=(x-3)(x-1)

b/ 3x²-5x+2=3x²-3x-2x+2=3x(x-1)-2(x-1)=(3x-2)(x-1)

a)   \(\left(x-7\right)\left(x+2\right)\)

b)  \(\left(2x-3\right)\left(x+2\right)\)

a) 3x² + 10x + 8

= 3x² + 6x + 4x + 8

= 3x.(x + 2) + 4.(x + 2)

= (x + 2).(3x + 4)

b) 2x² + x - 6

= 2x² + 4x - 3x - 6

= 2x.(x + 2) - 3.(x + 2)

= (x + 2).(2x - 3)

a)    \(\left(3x+4\right)\left(x+2\right)\)

b)    \(\left(2x-3\right)\left(x+2\right)\)

a) 3x² + 10x + 8

= 5x² - 2x² + 10x + 8

= (5x² + 10x) + (-2x² + 8)

= 5x(x + 2) - 2(x² - 4)

= 5x(x + 2) - 2(x + 2) (x - 2)

rùi bn làm tiếp nha mk bí!! 6575685685976978454254534565654567567567567563454

\(\left(x-1,5\right)^6+2.\left(1,5-x\right)^2=0\\ \Leftrightarrow\left(x-1,5\right)^6+2.\left(x-1,5\right)^2=0\\ \Leftrightarrow\left(x-1,5\right)^2.\left[\left(x-1,5\right)^4+2\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1,5\right)^2=0\\\left(x-1,5\right)^4+2\ge0\forall x\in R\end{matrix}\right.\\ \Leftrightarrow x=1,5\)

Vậy x=1,5

Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)

\(\Leftrightarrow x-1.5=0\)

hay x=1,5

a² - 4a + 3

= a² - a - 3a + 3

= a.(a - 1) - 3.(a - 1)

= (a - 1).(a - 3)

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

\(=x^8-x^7+x^5-x^4+x^2+x^7-x^6+x^4-x^3+x+x^6-x^5+x^3-x^2+1\)

\(=x^2\left(x^6-x^5+x^3-x^2+1\right)+x\left(x^6-x^5+x^3-x^2+1\right)+\left(x^6-x^5+x^3-x^2+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

\(x^8+x+1\)

\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)