d) \(ax^2-5x^2-ax+5x+a-5=\left(ax^2-ax+a\right)+\left(-5x^2+5x-5\right)\)
\(=a\left(x^2-x+1\right)-5\left(x^2-x+1\right)=\left(a-5\right)\left(x^2-x+1\right)\)
e) \(ax-bx-2cx-2a+2b+4c=x\left(a-b-2c\right)-2\left(a-b-2c\right)\)
\(=\left(x-2\right)\left(a-b-2c\right)\)
ax^2 - 5x^2-ax+5x+a-5
=x^2(a-5) -x(a-5)+(a-5)
=(a-5)(x^2-x+1)
ax-bx-2cx-2a+2b+4c
=x(a-b-c) -2(a-b-c)
=(x-2)(a-b-c)
Nhập ĐT nên k tiện gõ công thức, thông cảm
d, \(ax^2-5x^2-ax+5x+a-5\)
\(=\left(ax^2-ax+a\right)-\left(5x^2-5x+5\right)\)
\(=a\left(x^2-x+1\right)-5\left(x^2-x+1\right)\)
\(=\left(a-5\right)\left(x^2-x+1\right)\)
e, \(ax-bx-2cx-2a+2b+4c\)
\(=a\left(x-2\right)-b\left(x-2\right)-2c\left(x-2\right)\)
\(=\left(x-2\right)\left(a-b-2c\right)\)
f, \(\left(4x^2-7x-50\right)^2-16x^4-56x^3-49x^2\)
\(=\left(4x^2-7x-50\right)\left(4x^2-7x-50\right)-16x^4-56x^3-49x^2\)
\(=16x^4-28x^3-200x^2-28x^3+49x^2+350x-200x^2+350x+2500-16x^4-56x^3-49x^2\)
\(=-112x^3-400x^2+700x+2500\)
\(=-\left(112x^2+400x^2-700x-2500\right)\)
\(=-\left(112x^3-280x^2+680x^2-1700x+1000x-2500\right)\)
\(=-\left[112x^2\left(x-2,5\right)+680x\left(x-2,5\right)+1000\left(x-2,5\right)\right]\)
\(=-\left[\left(x-2,5\right)\left(112x^2+680x+1000\right)\right]\)
\(=-\left[\left(x-2,5\right)\left(112x^2+280x+400x+1000\right)\right]\)
\(=-\left\{\left(x-2,5\right)\left[112x\left(x+2,5\right)+400\left(x+2,5\right)\right]\right\}\)
\(=-\left[\left(x-2,5\right)\left(x+2,5\right)\left(112x+400\right)\right]\)
\(=-\left[16\left(x-2,5\right)\left(x+2,5\right)\left(7x+25\right)\right]\)
Chúc bạn học tốt!!!
Nguyễn Huy Tú Ace Legona Hồng Phúc Nguyễn Akai Haruma Nguyễn Thanh Hằng nguyen van tuan Mới vô Toshiro Kiyoshi TFBoys ... và những bạn giỏi toán vào đây giúp mik với
