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20 tháng 9 2020

A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)

\(=\frac{1}{x+3}-\frac{1}{x+34}\)

\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Rightarrow x=31\)

Vậy, x = 31 

20 tháng 9 2020

Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với    \(x,k\inℝ;x\ne0;x\ne-k\)

Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)

23 tháng 3 2017

Ta có:

\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)

\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)

\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)

\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)

\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)

b/ Thế vô rồi tính nhé

23 tháng 3 2017

Đoạn gần cuối thay y-x= 1 luôn 

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)

\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\)  giờ mới thay không biết đã tối giản chưa

20 tháng 7 2020

\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}+\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)

\(B=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}+\frac{x-1}{x+3}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)

\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\div\frac{x+2}{x+3}\)

\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-x-12}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)

\(B=\left(\frac{21+x^2-x-12+x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)

\(B=\frac{2x^2-5x+12}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{\left(x+2\right)}\)

\(B=\frac{2x^2-5x+12}{\left(x-3\right)\left(x+2\right)}\)

\(B=\frac{2x^2-5x+12}{x^2-x-6}\)

Đến đây là chịu ạ :(

27 tháng 11 2015

\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)

 

 

27 tháng 11 2015

1/ (x+1)(x+2) +1/ (x+2)(x+3) +1/ (x+3)(x+4) +1/ (x+4)(x+5)

=1/x+1 -1/x+2 +1/x+2 -1/x+3 +1/x+3 -1/x+4 +1/x+4 -1/x+5

=1/x+1 -1/x+5

=4/(x+1)(x+5)

16 tháng 12 2020

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)

16 tháng 12 2020

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}\)

\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)

\(=\frac{x-5-x}{x\left(x-5\right)}\)

\(=-\frac{5}{x\left(x-5\right)}\)