\(\frac{2.\left(-13\right).5.7.9}{\left(-3\right).4.\left(-5\right).26.14}\)
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HH
30 tháng 7 2018
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
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: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)