Áp dụng hằng đẳng thức đáng nhớ để thực hiện phép chia:
(125x3 + 1) : (5x + 1)
= [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1
= 25x2 – 5x + 1.
Cho mình hỏi làm sao lại ra (5x)2 – 5x + 1 ?
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(125x3 + 1) : (5x + 1)
= [(5x)3 + 1] : (5x + 1)
= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)
= (5x)2 – 5x + 1
= 25x2 – 5x + 1
Lời giải:
a) (x2 + 2xy + y2) : (x + y)
= (x + y)2 : (x + y)
= x + y
b) (125x3 + 1) : (5x + 1)
= [(5x)3 + 1] : (5x + 1)
= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)
= (5x)2 – 5x + 1
= 25x2 – 5x + 1
c) (x2 – 2xy + y2) : (y – x)
= (x – y)2 : [-(x – y)]
= -(x – y)
= y – x
Hoặc (x2 – 2xy + y2) : (y – x)
= (y2 – 2yx + x2) : (y – x)
= (y – x)2 : (y – x)
= y – x
a) (x^2+2xy+y^2) : (x+y)
=(x+y)2:(x+y)
=x+y
b) (125x^3+1) : (5x+1)
=(5x+1)(25x2-5x+1):(5x+1)
=25x2-5x+1
c) (x^2-2xy+y^2) : (y-x)
=(x-y)2:(y-x)
=-(x-y)2:(x-y)
=-(x-y)
=-x+y
\(\left(125x^3+1\right):\left(5x+1\right)\)
mà \(125x^3+1=\left(5x\right)^3+1=\left(5x+1\right)\left(25x^2-5x+1\right)\)
\(\Rightarrow\frac{\left(5x+1\right)\left(25x^2-5x+1\right)}{5x+1}=25x^2-5x+1\)
a,(x+2y)3 =x3+3.x2.2y+3.x.(2y)2+(2y)3
= x3+6x2y+12xy2+8y3
b, phần b tương tự dấu thay đổi một tí
c, (5x+1)(5x+1)= (5x+1)2
=25x2+10x+1
a) (x2 + 2xy + y2) : (x + y);
=(x+y)2:(x+y)
=x+y
b) (125x3 + 1) : (5x + 1);
=(5x+1)(25x2-5x+1):(5x+1)
=25x2-5x+1
c) (x2 – 2xy + y2) : (y – x).
=(x-y)2:(y-x)
=(y-x)2:(y-x)
=y-x
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
Bài giải:
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
(125x3+1):(5x+1)
= [(5x)3+1)]:(5x+1)
=(5x+1)(25x2-5x+1):(5x+1)
= 25x2-5x+1