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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
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: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
\(\frac{1}{2}+\left(\frac{16}{21}+\frac{27}{13}\right)-\left(\frac{14}{13}-\frac{5}{21}\right)\)
\(=\frac{1}{2}+\frac{16}{21}+\frac{27}{13}-\frac{14}{13}+\frac{5}{21}\)
\(=\left(\frac{16}{21}+\frac{5}{21}\right)+\left(\frac{27}{13}-\frac{14}{13}\right)+\frac{1}{2}\)
\(=1+1+\frac{1}{2}\)
\(=\frac{5}{2}\)
#)Giải :
\(\frac{1}{2}+\left(\frac{16}{21}+\frac{27}{13}\right)-\left(\frac{14}{13}-\frac{5}{21}\right)\)
\(=\frac{1}{2}+\frac{16}{21}+\frac{27}{13}-\frac{14}{13}+\frac{5}{21}\)
\(=\frac{1}{2}+\left(\frac{16}{21}+\frac{5}{21}\right)+\left(\frac{27}{13}-\frac{14}{13}\right)\)
\(=\frac{1}{2}+1+1\)
\(=2\frac{1}{2}=\frac{5}{2}\)