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đùa à bất kì số thập phân hữu hạn nào luỹ thừa lên cũng = 0 nên nó rất dễ giải:
a) \(\left[0,\left(3\right)\right]^2-\dfrac{81}{82}+2\)
= \(0-\dfrac{81}{82}+2\)
= \(\left(0+2\right)-\dfrac{81}{82}\)
= \(2-\dfrac{81}{82}\)
= \(\dfrac{83}{82}\)
b) \(3-\dfrac{1}{49}+\left[0,\left(142857\right)\right]^2\)
= \(3-\dfrac{1}{49}+0\)
= \(\left(3+0\right)-\dfrac{1}{49}\)
= \(3-\dfrac{1}{49}\)
= \(\dfrac{146}{49}\)
mình rút gọn luôn đó nhe 
a)
\(\left[0,\left(3\right)\right]^2-\dfrac{81}{82}+2\\ =\dfrac{1}{9}-\dfrac{81}{82}+2\\ =\dfrac{82}{738}-\dfrac{729}{738}+\dfrac{1479}{738}\\ =\dfrac{82-729+1479}{738}\\ =\dfrac{832}{738}\\ \approx1,13\)
b)
\(3-\dfrac{1}{49}+\dfrac{1}{7}\\ =\dfrac{147}{49}-\dfrac{1}{49}+\dfrac{7}{49}\\ =\dfrac{147-1+7}{49}\\ =\dfrac{153}{49}\\ \approx3,12\)
\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
\(\frac{7}{9}\)\(\times\)\(\sqrt{324}\)\(+\)\(\frac{1}{3}\)\(\times\)\(\sqrt{81}\)\(+\)\(\left(2014\right)^o\)\(-\)\(\left|-16\right|\)
\(=\)\(\frac{7}{9}\)\(\times\)18 + \(\frac{1}{3}\)\(\times\)9 + (2014)\(^0\)- |-16|
\(=\)14 + 3 + 1 - 16
\(=\)2
Bạn K cho mình nha
\(2^3+3.\left(\frac{2}{3}\right)^0-2+\left[\left(-2\right)^2:\frac{1}{2}\right]-8\)
đổi p/s \(\left(\frac{2}{3}\right)^0=1\)
xong tính trong ngoặc vuông,
r xử dụng tính chất phân phối
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
1.
a) \(3^3.9^{-1}\)
\(=27.\frac{1}{9}\)
\(=3.\)
b) \(25.5^{-1}.5^0\)
\(=25.\frac{1}{5}.1\)
\(=5.1\)
\(=5.\)
c) \(3^2.\frac{1}{243}.81^2.3^{-3}\)
\(=9.\frac{1}{243}.6561.\frac{1}{27}\)
\(=\frac{1}{27}.6561.\frac{1}{27}\)
\(=243.\frac{1}{27}\)
\(=9.\)
Chúc bạn học tốt!
a, \(3^3.9^{-1}\)
\(=27.\frac{1}{9}\)
\(=\frac{27}{9}=3\)
b, \(25.5^{-1}.5^0\)
\(=25.\frac{1}{5}.1\)
\(=\frac{25}{5}.1\)
\(=5.1\)
\(=5\)
c, \(3^2.\frac{1}{143}.81^2.3^{-3}\)
\(=9.\frac{1}{143}.6561.\frac{1}{3^3}\)
\(=9.\frac{1}{143}.6561.\frac{1}{27}\)
\(=9.\frac{1}{143}\left(6561.\frac{1}{27}\right)\)
\(=9.\frac{1}{143}.243\)
\(=\frac{9}{143}.243\)
\(=\frac{2187}{143}\)
Câu d tương tự các câu trên
nhanh nha 4h30 mik hc r