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\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)

15 tháng 12 2022

\(=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{9}{5}\cdot\dfrac{1}{9}+\dfrac{3}{4}}=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{1}{5}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{4}{25}+\dfrac{3}{5}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{16+60+75}{100}}=\dfrac{\sqrt{151}}{10}\)

15 tháng 12 2022

cảm ơn bạn

 

24 tháng 12 2022

\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

[3/7.4/15+1/3.(9^15)]^0.1/3.6^8/12^4

= 1.1/3.(2.3)^8/(3.4)^4

= 1/3.2^8.3^8/3^4.4^4

= 1/3.2^8.3^8/3^4.2^8

= 1/3.3^8/3^4

= 1/3.3^4=27

(dấu . là nhân nha)

30 tháng 12 2020

=1.1/3x6^8/6^8

=1/3.1

=1/3

gmae là dễ

 

22 tháng 6 2022

a) A=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114.
b) B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152.

13 tháng 7 2022

a) \mathrm{A}=\left[\dfrac{2}{7}\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right]:\left[\dfrac{2}{7}\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right]=\left(\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{1}{3}-\dfrac{2}{5}\right)=1 \dfrac{1}{4}.
b) \mathrm{B}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{3}+\dfrac{2}{7}\right)}{\dfrac{1}{5}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)-\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{3}\right)}{\left(\dfrac{1}{5}-\dfrac{1}{3}\right)\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=1 \dfrac{11}{52}