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1. \(\frac{25}{100}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{4}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\left(\frac{1}{4}+1-\frac{1}{5}\right)x=\frac{1}{5}\)
\(\Leftrightarrow\frac{21}{20}x=\frac{1}{5}\)
\(\Leftrightarrow x=\frac{1}{5}:\frac{21}{20}\)
\(\Leftrightarrow x=\frac{4}{21}\)
giúp mik nha chiều này 6:00 mik nộp rồi
ai nhanh mik sẽ k cho 3 k
\(2\frac{3}{5}x-\frac{1}{7}=1\frac{9}{35}\)
\(\frac{13}{5}x=\frac{44}{35}+\frac{1}{7}\)
\(\frac{13}{5}x=\frac{7}{5}\)
\(x=\frac{7}{5}:\frac{13}{5}\\ x=\frac{7}{13}\)
Mấy câu trên dễ , bạn có thể tự làm được
Chứng minh \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< 1\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1}-\frac{1}{10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{9}{10}\)
Lại có : \(\frac{9}{10}< 1\)
=> \(A< \frac{9}{10}< 1\)
=> \(A< 1\left(đpcm\right)\)
\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
bài khó nhất nhé
2. Ta có :
\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
cộng vào 48 phân số đầu với 1, trừ phân số cuối đi 48 ta được :
\(P=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\left(\frac{49}{1}-48\right)\)
\(P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(P=\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
\(P=50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}}{50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)}=\frac{1}{50}\)
1) \(\frac{2}{3}+x=-\frac{4}{5}\)
\(x=\left(-\frac{4}{5}\right)-\frac{2}{3}\)
\(x=-1\frac{7}{15}\)
Vậy \(x=-1\frac{7}{15}\)
2) \(\frac{2}{5}-x=-\frac{1}{3}\)
\(x=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
\(x=\frac{11}{15}\)
Vậy \(x=\frac{11}{15}\)
3) \(1-\frac{x}{3}=1\frac{1}{2}\)
\(\frac{x}{3}=1-1\frac{1}{2}\)
\(\frac{x}{3}=-\frac{1}{2}\)
\(\Rightarrow x=\frac{\left(-1\right)\cdot3}{2}\)
\(x=-1\frac{1}{2}\)
4) \(1-\left(\frac{2x}{3}+2\right)=-1\)
\(\frac{2x}{3}+2=1-\left(-1\right)\)
\(\frac{2x}{3}+2=2\)
\(\frac{2x}{3}=2-2\)
\(\frac{2x}{3}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)