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8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
a,\(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}...+\dfrac{1}{29+30}\)
\(A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{30}\)
\(A=\dfrac{1}{3}-\dfrac{1}{30}\)
\(A=\dfrac{9}{30}=\dfrac{3}{10}.\)
b, \(B=\dfrac{4}{7.11}+\dfrac{9}{11.20}+\dfrac{5}{20.25}\)
\(B=\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{25}\)
\(B=\dfrac{1}{7}-\dfrac{1}{25}\)
\(B=\dfrac{18}{175}\).
a/ \(\dfrac{x}{-\dfrac{5}{9}}=\dfrac{\dfrac{4}{3}}{-\dfrac{2}{5}}\)
\(\Leftrightarrow x.\left(\dfrac{-2}{5}\right)=\left(\dfrac{-5}{9}\right).\dfrac{4}{3}\)
\(\Leftrightarrow x.\left(-\dfrac{2}{5}\right)=\dfrac{-20}{27}\)
\(\Leftrightarrow x=\)\(\dfrac{-54}{100}\)
b/ tương tự
b/ \(\dfrac{7}{\dfrac{5}{3x}}=\dfrac{-\dfrac{4}{9}}{\dfrac{5}{6}}\)
\(\Leftrightarrow\dfrac{5}{3x}.\left(-\dfrac{4}{9}\right)=7.\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{5}{3x}.\left(-\dfrac{4}{9}\right)=\dfrac{35}{6}\)
\(\Leftrightarrow\dfrac{5}{3x}=\dfrac{-315}{24}\)
\(\Leftrightarrow3x\left(-315\right)=5.24\)
\(\Leftrightarrow3x\left(-315\right)=120\)
tự tính tiếp, đang lười
\(\dfrac{8}{9}+\dfrac{15}{23}+\dfrac{1}{9}+\dfrac{-15}{23}+\dfrac{1}{2}\)
\(=\left(\dfrac{8}{9}+\dfrac{1}{9}\right)+\left(\dfrac{-15}{23}+\dfrac{15}{23}\right)+\dfrac{1}{2}\)
\(=1+0+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
\(\dfrac{8}{9}+\dfrac{15}{23}+\dfrac{1}{9}+\dfrac{-15}{23}+\dfrac{1}{2}\)
=\((\dfrac{8}{9}+\dfrac{1}{9})\) +\((\dfrac{15}{23}+\dfrac{-15}{23})\) +\(\dfrac{1}{2}\)
= 1+ 0+\(\dfrac{1}{2}\)
= \(\dfrac{3}{2}\)
a/ \(\dfrac{\dfrac{-5}{12}}{\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|}=\dfrac{\dfrac{-4}{9}}{\dfrac{8}{15}}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\left(-\dfrac{5}{12}\right).\left(\dfrac{8}{15}\right)\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\dfrac{-2}{9}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\\dfrac{2}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\\dfrac{2}{3}x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy ....
a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}\)
\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)
\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)
\(=-\left(-\dfrac{173}{1287}\right)\)
\(=\dfrac{173}{1287}\)
c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{-49}{50}\)
a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)
câu b và c xem lại đề nha
Chúc bạn học tốt!!!
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
\(\dfrac{9}{9.19}+\dfrac{9}{19.29}+...+\dfrac{1}{89.99}\)
\(=9.\left(\dfrac{1}{9.19}+\dfrac{1}{19.29}+...+\dfrac{1}{79.89}\right)+\dfrac{1}{89.99}\)
\(=\dfrac{9}{10}.\left(\dfrac{10}{9.19}+\dfrac{10}{19.29}+...+\dfrac{10}{79.89}\right)+\dfrac{1}{89.99}\)
\(=\dfrac{9}{10}.\left(\dfrac{1}{9}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+...+\dfrac{1}{79}-\dfrac{1}{89}\right)+\left(\dfrac{1}{89}-\dfrac{1}{99}\right)\)
\(=\dfrac{9}{10}.\left(\dfrac{1}{9}-\dfrac{1}{89}\right)+\dfrac{10}{8811}\)
\(=\dfrac{9}{10}.\dfrac{80}{801}+\dfrac{10}{8811}\)
\(=\dfrac{8}{89}+\dfrac{10}{8811}\)
\(=\dfrac{802}{8811}\)
Đề là `9/[89.99]` thay vì là `1/[89.99]` chứ nhỉ? Vì nếu bạn để đề vậy thì không thấy quy luật là như thế nào.
*Sửa: `9/[9.19]+9/[19.29]+....+9/[89.99]`
`=9/10(10/[9.19]+10/[19.29]+....+10/[89.99])`
`=9/10(1/9-1/19+1/19-1/29+.....+1/89-1/99)`
`=9/10(1/9-1/99)`
`=9/10 . 10/99`
`=1/11`