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Ta có: \(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}=42^5:\left(2^3\cdot21^6\right)\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5\cdot21^5}{2^3\cdot21^5\cdot21}\)
\(\Leftrightarrow\left(x-2\right)\cdot\frac{4}{21}=\frac{4}{21}\)
\(\Leftrightarrow x-2=1\)
hay x=3
Vậy: x=3

\(\Leftrightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\frac{5}{14}\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Leftrightarrow x=2\)
\(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{7}{14}-\frac{2}{14}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{5}{14}.\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{21}:\frac{5}{14}\)
\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)

\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)
\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)
ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP

\(x\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)=\frac{4034}{5}\)
\(x.\frac{2}{5}=\frac{4034}{5}\)
\(x=\frac{4034}{5}:\frac{2}{5}\)
\(x=\frac{4034}{5}.\frac{5}{2}\)
x = 2017

Mình lỡ đánh nhầm 2 lần \(\frac{5}{14}\)nha :)) chỉ 1 lần thôi

Link này bạn:Câu hỏi của Hoàng Hà Nhi - Toán lớp 6 | Học trực tuyến
\(\dfrac{x-1}{12}+\dfrac{x-1}{20}+\dfrac{x-1}{30}+...+\dfrac{x-1}{72}=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{8.9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\dfrac{2}{9}=\dfrac{16}{9}\\ x-1=8\\ x=8+1\\ x=9\)

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x+2}{42} +\frac{x+2}{56}+\frac{x+2}{72}=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-2\right)\frac{4}{9}=\frac{16}{9}\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{3}{9}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\frac{2}{9}\)
\(x-2=\frac{16}{9}\cdot\frac{9}{2}\)
\(x-2=8\)
\(x=8+2\)
\(x=10\)
Vậy \(x=10\)
\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\)\(=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{2}{9}\right)=\frac{16}{9}\)
2(x-2)=16
x-2=8
x=10
Ta có: \(\frac{x}{20}+\frac{x}{30}+\frac{x}{42}=3\)
=>\(x\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=3\)
=>\(x\left(\frac14-\frac15+\frac15-\frac16+\frac16-\frac17\right)=3\)
=>\(x\left(\frac14-\frac17\right)=3\)
=>\(x\cdot\frac{3}{21}=3\)
=>\(x=3:\frac{3}{21}=21\)
\(\frac{x}{20}+\frac{x}{30}+\frac{x}{42}=\) \(3\)
\(\frac{21x}{420}+\frac{14x}{420}+\frac{10x}{420}=3\)
\(\frac{45x}{420}=3\)
\(45x=3\times420\)
\(45x=1260\)
\(x=1260:45\)
\(x=28\)