1 thuc hien phep tinh
a, 3x2y (3x – 4y); b, ; c, (3x + 5)( 3xy - 5x)
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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)
\(\frac{1}{100.99}-\frac{1}{99.98}-......-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(-\frac{1}{100.99}+\frac{1}{99.98}+...........+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(-\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+......+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(-\frac{1}{100}-1\right)\)
\(=\frac{1}{100}+1\)
\(=\frac{101}{100}\)
Đề bài là " Thực hiện phép tính " , thế phép tính nào ? Mk có thấy phép tính nào đâu !?
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.
a: 3x^2y(3x-4y)
=3x^2y*3x-3x^2y*4y
=9x^3y-12x^2y^2
c: (3x+5)(3xy-5x)
=9x^2y-15x^2+15xy-25x