=>(x-\(\sqrt{5}\))²

=>(x-\(\sqrt{5}\)) (x-\(\sqrt{5}\))

(x + 2y)² - (x - y)²

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

= 3y(2x + y)

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

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

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

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

    \(\left(x^2-5\right)^2+144\)

\(=x^4-10x^2+25+144\)

\(=x^4-10x^2+169\)

\(=x^4+26x^2+169-36x^2\)

\(=\left(x^2+13\right)^2-\left(6x\right)^2\)

\(=\left(x^2-6x+13\right)\left(x^2+6x+13\right)\)

Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

Bài 1;

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

2) \(6x^2+12xy+6y^2=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(2y^3+8y^3+8y=10y^3+8y=2y\left(5y^2+4\right)\)

4) \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

1) \(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

2) \(8x^2y-18y=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\)

3) \(24x^3-3=3\left(8x^3-1\right)=3\left(2x-1\right)\left(4x^2+2x+1\right)\)

Bài 3:

1) \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y\right]=5\left(x-y+1\right)\left(x+y+1\right)\)

2) \(3x^3-6x^2+3x-12xy^2=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=3x\left(x-2y-1\right)\left(x+2y-1\right)\)

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

4) \(2x^3-2xy^2-8x^2+8xy=2x\left(x^2-y^2-4x+4y\right)=2x\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]=2x\left(x-y\right)\left(x+y-4\right)\)

\(A=5x^3-125x=5x\left(x-5\right)\left(x+5\right)\)

\(B=x^3-8+\left(x-2\right)\left(5x+4\right)\)

\(=\left(x-2\right)\left(x^2+2x+4+5x+4\right)\)

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

 [ (x+2)^2 + 5]^2 +3x^3 + 14x^2 + 24x 
= (x+2)^4 + 8[(x+2)]^2 + 16 + (3x^2).(x+2) + 8x(x+2) +8x 
=(x+2).[(x+2)^3 +8(x+2) + 3x^2 +8x +8 ] 
=(x+2).[x^3 + 9x^2 + 28x +32 ] 
=(x+2).(x+4).(x^2 +5x +8)

các bạn ơi làm ơn giải hộ mình bài này mình đang cần rất rất gấp

Bạn kiểm tra lại đề đi nhé (lớp 8 chưa học số vô tỉ, căn)

Nếu đề bạn đúng thì dùng công thức lớp 9 (cái này bạn tự tìm hiểu chứ mình cũng không nhớ).

81x⁴+4

=(9x²)²+2²

=(9x²)²+36x²+2²-36x²

=(9x²+2)²-(6x)²

=(9x²+2-6x)(9x²+2+6x)

8: \(=\left(x-2y\right)\cdot x\cdot\left(x+3\right)\)

9: \(=\left(5x+2\right)\left(x-3\right)-x\left(x-3\right)\)

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

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

3: \(=2\left(x+2\right)\left(25x-15-x\right)\)

\(=2\left(x+2\right)\left(24x-15\right)\)

=6(x+2)(8x-5)