\(a,=\left(x+4\right)^2\\ b,=\left(x-6\right)^2\\ c,=-\left(4x^2-4x+1\right)=-\left(2x-1\right)^2\\ d,=\left(x-1\right)^3\)

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

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

c. \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)

d. \(x^3-3x^2+3x-1\)

\(=x^3-1^3-3x^2+3x\)

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

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

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

e. \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

g. \(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)

a. (a² - b²)² - (a² + b²)²

= (a² - b² - a² - b²)(a² - b² + a² + b²)

= -2b² . 2a²

b. a⁶ - b⁶

<=> (a³)² - (b³)²

<=> (a³ - b³)(a³ + b³)

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

\(b,a^6-b^6=a^2\left(a^3-b^3\right)=a^2\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(c,-4x^2+9y^2=\left(3y-2x\right)\left(3y+2x\right)\\ d,\left(x+1\right)^3-\left(2-x\right)^3\\ =\left(x+1-2+x\right)\left[\left(x+1\right)^2+\left(x+1\right)\left(2-x\right)+\left(2-x\right)^2\right]\\ =\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)\\ =\left(2x-1\right)\left(x^2-x+7\right)\)

\(e,8+\left(4x-3\right)^3\\ =\left(8+4x-3\right)\left[64-8\left(4x-3\right)+\left(4x-3\right)^2\right]\\ =\left(4x+5\right)\left(64-32x+24+16x^2-24x+9\right)\\ =\left(4x+5\right)\left(16x^2-56x+97\right)\)

\(g,81-\left(9-x^2\right)^2\\ =\left(9-9+x^2\right)\left(9+9-x^2\right)\\ =x^2\left(18-x^2\right)\left[=x^2\left(\sqrt{18}-x\right)\left(\sqrt{18}+x\right)\right]\)

Chỗ trong ngoặc nếu bạn chưa học căn thì ko cần ghi nha

a: Ta có: \(\left(a^2-b^2\right)^2-\left(a^2+b^2\right)^2\)

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

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

b: \(a^6-b^6\)

\(=\left(a^3-b^3\right)\left(a^3+b^3\right)\)

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

d: \(\left(x+1\right)^3+\left(x-2\right)^3\)

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

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

\(=\left(2x-1\right)\left(x^2-x+7\right)\)

1) ( 4x + 1 )² + ( 4x - 1 )² - 2( 4x + 1 ).( 4x - 1 )

= ( 4x + 1 - 4x - 1 )²

= 2²

= 4

2) 4x² - 9 + ( 2x + 3 )

= ( 2x )² - 3² + ( 2x + 3 )

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

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

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

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

1, (4x+1)^2 + (4x-1)^2 - 2(4x+1)(4x-1)

=[(4x+1)-(4x-1)]^2

=(4x+1-4x+1)^2

=2^2

=4

2, 4x^2 - 9 +(2x+3)

=(4x^2 - 9)+(2x+3)

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

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

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

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

=.= hok tốt!!

B1 :

a) (2x - 1)²

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

\(=\left(x^3+2x^2\right)+\left(2x^2+4x\right)-16y^2\)

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

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

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

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

\(=\left[\sqrt{x}.\left(x+2\right)\right]^2-4y^2\)

\(=\left[\sqrt{x}.\left(x+2\right)-4y\right].\left[\sqrt{x}.\left(x+2\right)+4y\right]\)

Tham khảo nhé~

nếu đưa vô căn phải có điều kiện là x > 0

\(x^3+4x^2+4x-16y^2=x\left(x+2\right)^2-\left(4y\right)^2\)

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

1: \(4x^2+4x+1=\left(2x+1\right)^2\)

2: \(x^2-20x+100=\left(x-10\right)^2\)

3: \(y^4-14y^2+49=\left(y^2-7\right)^2\)

4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

a) \(=\left(2x-1\right)^2\)

b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)

a. \(4x^2-4x+1=\left(2x\right)^2-2x.2.1+1^2=\left(2x-1\right)^2\)

b. \(xy^2-x^3+2x^2-x=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x^2-2x+1\right)\right]=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)