Mình sửa: Bài 1
2)x²+3x-15

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

kết quả thôi nha

bài 1 : a, sai đề . đề đúng là x³ + x² + 2² = x³ + 2x²- x²+ 4

= x² ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 ) ( x² - x + 2 )

b, xy -y² -x -y = ( xy -x ) + ( y² - y ) = x ( y -1) + y ( y -1) = ( y -1 ) ( x +y )

ôi mai dê

mấy bài này max dễ bn đăng từng phần 1 mk lm cho

c,x^4-5x^2+4=x^4-4x^2-x^2+4=(x^2-4)(x^2-1)=(x-1)(x+1)(x-2)(x+2)

e,x^4-3x^3+x^2+3x-2=x^4-x^3-2x^3+2x^2-x^2+x+2x-2=(x-1)(x^3-2x^2-x+2)

Đến đây lấy máy tính bấm Mode*3+1+>+3 rồi tìm nghiệm

Các câu khác cũng máy tính đi

Bài 1 : 

a) \(x^4-4x^2-4x-1\)

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

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

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

b) \(x^2+2x-15\)

\(=x^2+2x+1-16\)

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

\(=\left(x+1+4\right)\left(x+1-4\right)=\left(x+5\right)\left(x-3\right)\)

c) \(x^3y-2x^2y^2+5xy\)

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

B2:

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

\(=2\left(x^2-2x+1\right)-\left(4x^2-9\right)\)

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

\(=-2x^2-4x+11\)

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

\(=\left(x+3-x+3\right)^2=6^2=36\)

c) \(4\left(x-1\right)\left(x+3\right)+5\left(2x+1\right)^2-2\left(5-3x\right)^2\)

\(=4\left(x^2+2x-3\right)+5\left(4x^2+4x+1\right)-2\left(9x^2-30x+25\right)\)

\(=4x^2+8x-12+20x^2+20x+5-18x^2+60x-50\)

\(=6x^2+88x-57\)

a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)

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

\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)

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

\(=\left(x-3\right)\left(2x^2-11x\right)\)

\(=x\left(x-3\right)\left(2x-11\right)\)

b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)

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

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

\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)

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

\(=\left(x-2y+5\right)\left(x+2y+1\right)\)

a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)

\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)

b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)

 \(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)