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â, -4/9(7/15+8/15)=-4/9
b,-5/4(16/25+9/25)=-5/4
,.....
dài quá mik làm ko hết
hok tốt
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) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
Câu 3 : \(2+4+6+.........+2n=156\)
\(\Leftrightarrow2\left(1+2+3+.....+n\right)=156\)
\(\Leftrightarrow1+2+3+.........+n=78\)
\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=78\)\(\Leftrightarrow n\left(n+1\right)=156=12.13\)\(\Leftrightarrow n=12\)
Vậy \(n=12\)
Mk chỉ biết câu a thôi
a) \(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}\)
= \(\frac{5\cdot\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}{15\cdot\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}\)
= \(\frac{5}{15}\)
= \(\frac{1}{3}\)
Chúc bạn học tốt
=1/1-1/2+1/2-1/3+1/3-1/4+.........+1/1999-1/2000
=1/1-1/2000
=1999/2000<3/4
\(\frac{-7}{11}.\frac{11}{19}+\frac{-7}{11}.\frac{8}{19}+\frac{-4}{11}\)
\(=\frac{-7}{11}.\left(\frac{11}{19}+\frac{8}{19}\right)+\frac{-4}{11}\)
\(=\frac{-7}{11}.1+\frac{-4}{11}\)
\(=\frac{-7}{11}+\frac{-4}{11}=\frac{-11}{11}=-1\)
~ Hok tốt ~
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow B=1-\frac{1}{2019}\)
\(\Rightarrow B=\frac{2018}{2019}\)