Question

giúgiúp tui với ạ trước 10h nha tui đang cần gấp

Answer 1

Bài trên:

\(16x^3y+0,25yz^3=\dfrac{1}{4}y\left(64x^3+z^3\right)=\dfrac{1}{4}y\left[\left(4x\right)^3+z^3\right]\\ =\dfrac{1}{4}y\left[\left(4x+z\right)\left(16x^2-4xz+z^2\right)\right]\\ ----\\ x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\\ -----\\ a^3+a^2b-ab^2-b^3=\left(a^3-b^3\right)+\left(a^2b-ab^2\right)\\ =\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)=\left(a-b\right)\left(a^2+2ab+b^2\right)=\left(a-b\right)\left(a+b\right)^2\)

Answer 2

Bài trên

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

Answer 3

Bài trên:

\(3a-3b+a^2-2ab+b^2\\ =3\left(a-b\right)+\left(a-b\right)^2\\ =\left(3+a-b\right)\left(a-b\right)\\ ----\\ a^2-b^2-4a+4b\\ =\left(a-b\right)\left(a+b\right)-4\left(a-b\right)\\ =\left(a-b\right)\left(a+b-4\right)\\ ----\\ a^3-b^3-3a+3b\\ =\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\\ =\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ x^3+3x^2-3x-1\\ =x^3-x^2+4x^2-4x+x-1\\ =x^2\left(x-1\right)+4x\left(x-1\right)+\left(x-1\right)\\ =\left(x-1\right)\left(x^2+4x+1\right)\\ =\left(x-1\right)\left(x+2-\sqrt{3}\right)\left(x+2+\sqrt{3}\right)\\ \\---\\ x^3-3x^2-3x+1\\ =x^3+x^2-4x^2-4x+x+1\\ =x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)\\ =\left(x^2-4x+1\right)\left(x+1\right)\\ =\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)\left(x+1\right)\)

Answer 4

Bài 9: Ý 1

\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^4+2x^3+x^2\right)+4x^2+4x-12\\ =x^4+2x^3+5x^2+4x-12\\ =x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\\ =x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\\ =\left(x-1\right)\left(x^3+3x^2+8x+12\right)\\ =\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\\ =\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\\=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

Answer 5

Bài 9:

Ý 9

\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\\ =\left(4x^2+60x+200\right)\left(x^2+18x+72\right)-3x^2\\ =4x^4+132x^3+1568x^2+7920x+14400-3x^2\\ =4x^4+132x^3+1565x^2+7920x+14400\\ =4x^4+32x^3+100x^3+800x^2+765x^2+6120x+1800x+14400\\ =4x^3\left(x+8\right)+100x^2\left(x+8\right)+765x\left(x+8\right)+1800\left(x+8\right)\\ =\left(x+8\right)\left(4x^3+100x^2+765x+1800\right)\\ =\left(x+8\right)\left(4x^3+30x^2+70x^2+525x+240x+1800\right)\\ =\left(x+8\right)\left[4x^2\left(x+7,5\right)+70x\left(x+7,5\right)+240\left(x+7,5\right)\right]\\ =\left(x+8\right)\left(x+7,5\right)\left(4x^2-70x+240\right)\)

Answer 6

Bài 9 ý 8:

\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12\\ =\left(x^4+2x^3+x^2\right)+4x^2+4x-12\\ =x^4+2x^3+5x^2+4x-12\\ =x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\\ =x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\\ =\left(x-1\right)\left(x^3+3x^2+8x+12\right)\\ =\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\\ =\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\\ =\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)