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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) \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)
= \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\)
= \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
= \(\frac{1}{2}.\frac{4}{15}\)
= \(\frac{2}{15}\)
b)=(2/3 +2/7 - 2/28)/(-3/3 -3/7 + 3/28)
=[2(1/3+1/7-1/28)]/[(-3)(1/3+1/7-1/28)]
=2/-3
=-2/3
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) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
TÌM X
a,\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
b, \(\left(x-\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
Bài làm
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\) b) \(\left(x-\frac{1}{2}\right)^2=\frac{4}{25}\)
=> \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\) => \(\left(x-\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)
=> \(x-\frac{1}{2}=\frac{1}{3}\) => \(x-\frac{1}{2}=\frac{2}{5}\)
\(x=\frac{1}{3}+\frac{1}{2}\) \(x=\frac{2}{5}+\frac{1}{2}\)
\(x=\frac{2}{6}+\frac{3}{6}\) \(x=\frac{4}{10}+\frac{5}{10}\)
\(x=\frac{5}{6}\) \(x=\frac{9}{10}\)
Vậy \(x=\frac{5}{6}\) Vậy \(x=\frac{9}{10}\)
# Chúc bạn học tốt #
\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)\)
\(taco:\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}=\frac{-7}{6}:\frac{-1}{4}=\frac{14}{3}\)
\(\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)=\left(\frac{-5}{6}+\frac{-16}{3}\right)\cdot\left(-14\right)=\frac{-37}{6}\cdot\left(-14\right)=\frac{259}{3}\)
TU DO \(=>X=\frac{14}{3};\frac{15}{3};,,,;\frac{259}{3}\)
CHUC BAN HOC TOT :))