\(\frac{-8}{10}\).\(10\frac{4}{7}\)+\(2\frac{3}{7}\).\(\frac{-8}{13}\)+\(\frac{3}{5}\)(tính)
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\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)
\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)
\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)
\(A=\frac{154}{11}+\frac{8}{5}\)
\(\Rightarrow A=\frac{78}{5}\)
\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)
\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)
\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)
\(B=\frac{27}{9}+\frac{-72}{13}\)
\(B=3+\frac{-72}{13}\)
\(\Rightarrow B=\frac{-33}{13}\)
P/s: Hoq chắc :v
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
Câu a:
(4\(\frac{5}{37}\) - 3\(\frac45\) + 8\(\frac{15}{29}\)) - (3\(\frac{5}{37}\) - 6\(\frac{14}{29}\))
= 4\(\frac{5}{37}\) - 3\(\frac45\) + 8\(\frac{15}{29}\) - 3\(\frac{15}{37}\) + 6\(\frac{14}{29}\)
= (4 - 3) + (\(\frac{5}{37}-\frac{5}{37}\)) + (8 + 6) + (\(\frac{15}{29}\) + \(\frac{14}{29}\)) - 3 - \(\frac45\)
= 1 + 0 + 14 + 1 - 3 - \(\frac45\)
= 1 + 14 + 1 - 3 - \(\frac45\)
= 15 + 1 - 3 - \(\frac45\)
= 16 - 3 - \(\frac45\)
= 13 - \(\frac45\)
= \(\frac{61}{5}\)
a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)
\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)
b)\(5+\dfrac{3}{5}=\dfrac{25}{5}+\dfrac{3}{5}=\dfrac{28}{5};10-\dfrac{9}{16}=\dfrac{160}{16}-\dfrac{9}{16}=\dfrac{151}{16}\)
\(\dfrac{2}{3}-\left(\dfrac{1}{6}+\dfrac{1}{8}\right)=\dfrac{2}{3}-\left(\dfrac{8}{48}+\dfrac{6}{48}\right)=\dfrac{2}{3}-\dfrac{14}{48}=\dfrac{32}{48}-\dfrac{14}{48}=\dfrac{3}{8}\)
Có:495.810/147.49.413=710.230/233.79=7/8 7/10-7/12+7/5 / 8/10-8/12+8/5=7(1/10-1/12+1/5) / 8(1/10-1/12+1/5)=7/8 =>A=7/8-7/8=0 xin lỗi nha mik ghi hơi khó hiểu chút vì mik mới dùng online math.HOK TỐT



\(\frac{-8}{10}.\frac{74}{7}+\frac{17}{7}.\frac{-8}{13}+\frac{3}{5}=\frac{-8}{7}.\frac{37}{5}+\frac{-8}{7}.\frac{17}{13}+\frac{3}{5}\)
\(=\frac{-8}{7}.\left(\frac{37}{5}+\frac{17}{13}\right)+\frac{3}{5}=\frac{-8}{7}.\frac{566}{65}+\frac{3}{5}\)
\(=\frac{-4528}{455}+\frac{273}{455}=\frac{-4255}{455}=\frac{851}{91}\)
851/91