Ta có:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)

\(=1-\frac{2n+1}{\left(n+1\right)^2}\)

Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)

SAI RỒI ĐÁP ÁN LÀ N^2/(N+1)^2

Ta có:

\(2n+1=\left(n^2+2n+1\right)-n^2=\left(n+1\right)^2-n^2\Rightarrow\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)

Thay vào ta rút gọn được các số hạng của A, cuối cùng được:

\(A=1-\frac{1}{\left(n+1\right)^2}\)

a, \(x^4+2013x^2+2012x+2013\)

\(=x^4+2013x^2-x+2013x+2013\)

\(=\left(x^4-x\right)+\left(2013x^2+2013x+2013\right)\)

\(=x\left(x^3-1\right)+2013\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2013\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left\{x\left(x-1\right)+2013\right\}\)

\(=\left(x^2+x+1\right)\left(x^2-x+2013\right)\)

bạn đợi mình tí nhé, khoảng 15 phút sau mình gửi câu trả lời :>>

\(\text{Câu 1 thì bạn tự làm chịu khó tí:}y^2-4x^2=\left(y+2x\right)\left(y-2x\right)\)

\(\text{Bài 2:}ab\left(b-a\right)+bc\left(c-b\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left(b-c\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left[\left(b-a\right)+\left(a-c\right)\right]+ac\left(a-c\right)\)

\(=\left(ab-bc\right)\left(b-a\right)+\left(ac-bc\right)\left(a-c\right)\)

\(=b\left(a-c\right)\left(b-a\right)+c\left(a-b\right)\left(a-c\right)=\left(b-c\right)\left(b-a\right)\left(a-c\right)\)

1a)  Ta có:A = \(\left(\frac{2}{2x-y}+\frac{6y}{y^2-4x^2}-\frac{4}{2x+y}\right):\left(1+\frac{4x^2+y^2}{4x^2-y^2}\right)\)

A = \(\left(\frac{2\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}-\frac{6y}{\left(2x-y\right)\left(2x+y\right)}-\frac{4\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)}\right):\left(\frac{4x^2-y^2+4x^2+y^2}{4x^2-y^2}\right)\)

A = \(\frac{4x+2y-6y-8x+4y}{\left(2x+y\right)\left(2x-y\right)}\cdot\frac{\left(2x+y\right)\left(2x-y\right)}{8x^2}\)

A = \(\frac{-4x}{8x^2}=-\frac{1}{2x}\)

b) Với x = 0,25; y = 2015 => A = -1/2.0,25 = -1/0,5 = -2

2. Ta có: ab(b - a) - bc(b - c) - ac(c - a)

= ab² - a²b - b²c + bc² + ac(a - c)

= b²(a - c) - b(a - c)(a + c) + ac(a - c)

= (b² - ab - bc + ac)(a - c)

= [b(b - a) - c(b - a)](a - c)

= (b - a)(b - c)(a - c)

Bài làm

Như đã nhắn là mình sẽ làm theo quan điểm của mình là 5/(x^2 - 1) nha

\(A=\left[\frac{3\left(x+2\right)}{2x^3+2x+2x^2+2}+\frac{2x^2-x-10}{2x^3-2-2x^2+2x}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2x+2}-\frac{3}{2x-2}\right]\)

\(A=\left[\frac{3\left(x+2\right)}{2x^2\left(x+1\right)+2\left(x+1\right)}+\frac{2x^2+4x-5x-10}{\left(2x^3-2x^2\right)+\left(2x-2\right)}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{2x\left(x+2\right)-5\left(x+2\right)}{2x^2\left(x-1\right)+2\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{\left(2x-5\right)\left(x+2\right)}{\left(2x^2+2\right)\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}+\frac{\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3\left(x-1\right)}{2\left(x^2-1\right)}-\frac{3\left(x+1\right)}{2\left(x^2-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)+\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10}{2\left(x^2-1\right)}+\frac{3x-3}{2\left(x^2-1\right)}-\frac{3x+3}{2\left(x^2-1\right)}\right]\)

\(A=\left[\frac{\left(x+2\right)\left[3x-3+\left(2x-5\right)\left(x+1\right)\right]}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10+3x-3-3x-3}{2\left(x^2-1\right)}\right]\)

\(A=\left[\frac{\left(x+2\right)\left(3x-3+2x^2+2x-5x-5\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\frac{4}{2\left(x^2-1\right)}\)

\(A=\frac{\left(x+2\right)\left(2x^2-8\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\cdot\frac{\left(x^2-1\right)}{2}\)

\(A=\frac{\left(x+2\right)2\left(x^2-4\right)}{2\left(2x^2+2\right)}\)

\(A=\frac{2\left(x+2\right)\left(x-2\right)\left(x+2\right)}{4\left(x^2+1\right)}\)

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

:>>> Chả biết đúng không nữa nhưng số to quá :>> 

\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(A=1-\dfrac{1}{n^2+2n+1}\)

\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

\(2a,\left(6x+7\right)\left(2x-3\right)-\left(4x+1\right)\left(3x-\frac{7}{4}\right)\)

\(=12x^2-18x+14x-21-12x^2+7x-3x+\frac{7}{4}\)

\(=-21+\frac{7}{4}\)chứng tỏ biểu thức ko phụ thuộc vào biến x

3, Đặt 2n+1=a^2; 3n+1=b^2=>a^2+b^2=5n+2 chia 5 dư 2

Mà số chính phương chia 5 chỉ có thể dư 0,1,4=>a^2 chia 5 dư 1, b^2 chia 5 dư 1=>n chia hết cho 5(1)

Tương tự ta có b^2-a^2=n

Vì số chính phươn lẻ chia 8 dư 1=>a^2 chia 8 dư 1 hay 2n chia hết cho 8=> n chia hết cho 4=> n chẵn

Vì n chẵn => b^2= 3n+1 lẻ => b^2 chia 8 dư 1

Do đó b^2-a^2 chia hết cho 8 hay n chia hết cho 8(2)

Từ (1) và (2)=> n chia hết cho 40