x.x-2.x-1-y.y

\(\) \(x^2-2x-1-y^2=(x^2-2x+1)-2+y^2=(x-1)^2+y^2-2=((x-1)-y)((x-1)+y)-2=(x-1-y)(x+1+y)+2\)

     \(-16x^2+8xy-y^2+49\)

\(=49-\left(16x^2-8xy+y^2\right)\)

\(=7^2-\left(4x-y\right)^2\)

\(=\left(7-4x+y\right)\left(7+4x-y\right)\)

\(16x^2+y^2+4y-16y-8xy\)

\(=\left(16x^2-8xy+y^2\right)+4y-16y\)

\(=\left(4x+y\right)^2-12y\)

\(=\left(4x+y-\sqrt{12y}\right)\left(4x+y-\sqrt{12y}\right)\)

P/S : Sai thì thôi nha!

Kimetsu no YaibaNếu y âm thì căn thức vô nghĩa

x³ - x² + 16x - 16 = x²(x - 1) + 16(x - 1) = (x² + 16)(x - 1)

x^3-x^2+16x-16 = (x^3-x^2)+(16x-16) = x^2.(x-1) + 16.(x-1) = (x-1)(x^2+16)

SAI ĐỀ CÂU A NHÉ

CÂU B Ở CÂU HỎI TƯƠNG TỰ

\(x^4-4x^3+8x^2-16x+16 \)

\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)

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

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

A, (x+2)² - x² +2x -1

= (x+2)² -(x² - 2x +1)

= (x+2)² - (x -2)²

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

=2x.2 =4x

B, 16x² -y² 

= (4x)² - y²

= (4x - y).(4x + y)

Tk mình với bạn ơi. Đúng rồi nhé!!

CHÚC BẠN HỌC TỐT ✓✓

1, \(\left(x+2\right)^2-x^2+2x-1\)

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

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

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

\(16x^4+8x^2+1-8x^2\)

\(=\left(4x^2+1\right)^2-8x^2\)

\(=\left(4x^2+1-x\sqrt{8}\right)\left(4x^2+1+x\sqrt{8}\right)\)