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

\(=\left(3x^2-4x+5\right).2x^2-4\left(3x^2-4x+5\right)-6x^4+8x^3-16x\)

\(=6x^4-8x^3+10x^2-12x^2+16x-20-6x^4+8x^3-16x\)

\(=\left(6x^4-6x^4\right)+\left(8x^3-8x^3\right)-\left(12x^2-10x^2\right)+\left(16x-16x\right)-20\)

\(=-2x^2-20\)

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

\(=2\left(1-3x+x^2\right)-4x\left(1-3x+x^2\right)+4x^3+10x\)

\(=2-6x+2x^2-4x+12x^2-4x^3+4x^3+10x\)

\(=\left(4x^3-4x^3\right)+\left(12x^2+2x^2\right)+\left(10x-6x-4x\right)+2\)

\(=14x^2+2\)

\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)

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

\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)

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

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

\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)

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

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

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

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

\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)

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

\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)

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

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

Số lớn nhất có một chữ số là 9 nên thương là 9

Vì số chia là 8 nên số dư lớn nhất là 7

Số bị chia là: 8 x 9 + 7 = 79

Vậy Nam thực hiện phép chia 79 : 8 = 9 (dư 7)

giai ro rang

 (-1)+2-3+4-....-99+100 = 50

     \(\frac{1}{100.99}-\frac{1}{99.98}-......-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=-\left(-\frac{1}{100.99}+\frac{1}{99.98}+...........+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=-\left(-\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+......+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

\(=-\left(-\frac{1}{100}-1\right)\)

\(=\frac{1}{100}+1\)

\(=\frac{101}{100}\)

Đề bài là " Thực hiện phép tính " , thế phép tính nào ? Mk có thấy phép tính nào đâu !?

phép tính đâu

a)187 + [923 - (923 + 887)]

= 187 + [923 - 923 - 887]

= 187 + 923 - 923 - 887

= 187 + 0 - 887

= 187 - 887

= -700

b)39 . 187 + 30² - 78 . 39

= 39 . 187 + 900 - 78 . 39

= 39 . (187 - 78) + 900

= 39 . 109 + 900

= 4251 + 900

= 5151

a, 187+[923-(923+887)]

=187+[923-923-887]

=187-887

=-700

b, 39.187+30²-78.39

=39.(187-78)+900

=39.109+900

=4251+900

=5151

Câu b bạn xem lại đầu bài là 78 hay 87 nha. Là 87 sẽ đúng hơn đó.

=[x²-1].[x+2]

x³-x+2x²-2