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mk làm bài 2 trước nhé
\(\frac{x+2}{2}=\frac{72}{x+2}\)
\(=>\left(x+2\right)^2=72.2=144=12^2\)
\(=>x+2=12\)
\(=>x=12-2=10\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(B=\frac{1^2}{2}.\frac{2^2}{6}.\frac{3^2}{12}.....\frac{9^2}{90}=\frac{1^2.2^2.3^2.....9^2}{\left(1.2\right).\left(2.3\right).\left(3.4\right).....\left(9.10\right)}\)
\(=\frac{1^2.2^2.3^2.....9^2}{1.2^2.3^2.....9^2.10}=\frac{1}{10}\)
Bài 4:
\(50B=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}\)
\(50B=\frac{1+99}{1.99}+\frac{3+97}{3.97}+...+\frac{99+1}{49.51}\)
\(50B=1+\frac{1}{99}+\frac{1}{97}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{51}=A\)
\(\Rightarrow\frac{A}{B}=50\).
\(a,\frac{2}{3}\cdot x-\frac{4}{7}=\frac{1}{8}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{1}{8}+\frac{4}{7}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{7}{56}+\frac{32}{56}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{39}{56}\)
\(\Leftrightarrow x=\frac{39}{56}:\frac{2}{3}=\frac{39}{56}\cdot\frac{3}{2}=\frac{39\cdot3}{56\cdot2}=\frac{117}{112}\)
\(b,\frac{2}{7}-\frac{8}{9}\cdot x=\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{2}{7}-\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{6}{21}-\frac{14}{21}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{-8}{21}\)
\(\Leftrightarrow x=\frac{-8}{21}:\frac{8}{9}=\frac{-8}{21}\cdot\frac{9}{8}=\frac{-8\cdot9}{21\cdot8}=\frac{-1\cdot3}{7\cdot1}=\frac{-3}{7}\)
Làm nốt hai bài cuối đi nhé
Study well >_<
Mk k chép lại đề bài nha
a)\(\frac{2}{3}.x=\frac{1}{8}+\frac{4}{7}\)
\(\frac{2}{3}.x=\frac{7}{56}+\frac{32}{56}\)
\(\frac{2}{3}.x=\frac{39}{56}\)
\(x=\frac{39}{56}:\frac{2}{3}\)
\(x=\frac{39}{56}.\frac{3}{2}\)
\(x=\frac{117}{112}\)
Mk sợ sai lém!!!
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
a)(1,5.x+3/7):1,5-1,5=1 b)(2+x+4/7+3/7):1,4-5/7=-2
1,5.x:1,5+3/7:3/2=1+1,5 (2+x+1):1,4-5/7=-2
x+ 3/7x 2/3=2,5 (3+x):1,4=-2+5
x+2/7=2,5 (3+x):1,4=3
x+2/7=5/2 3+x=3x1,4
x=5/2-2/7 3+x=4,2
x=31/14 x=4,2-3=1,2