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làm bài 3 BĐT
theo bảng xét dấu
còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2
bài 1.2 (tức bài 2 ở trên )làm a,b,c,d
\còn bài 2( tức bài 2 ở trên) làm hết
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6\cdot\dfrac{5}{3}\cdot\dfrac{3}{7}:\left(\dfrac{7^3}{5^4}\right)^{-2}\)
\(=\left(\dfrac{5}{7}\right)^6\cdot\dfrac{5}{7}\cdot\left(\dfrac{5}{7}\right)^6\cdot5^2\)
\(=\left(\dfrac{5}{7}\right)^{13}\cdot5^2\)
c: \(=5^4\cdot2.5^{-5}\cdot125\cdot0.04\)
\(=5^4\cdot5\cdot\left(\dfrac{5}{2}\right)^{-5}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
1.
a) x : \(\left(\dfrac{3}{4}\right)^3\) =\(\left(\dfrac{3}{4}\right)^3\)
x = \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{3}{4}\right)^3\)
x = \(\dfrac{3}{4}^{3+3}\)
x = \(\dfrac{3}{4}^6\)
x = \(\dfrac{729}{4096}\)
b) \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
x = \(\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\)
x = \(\dfrac{2}{5}^{8-5}\)
x = \(\dfrac{2}{5}^3\)
x = \(\dfrac{8}{5}\)
2.
(0,36)\(^8\) \([\left(0,6\right)^3]^8\) = (0,6)\(^{3.8}\) = ( 0,6)\(^{24}\)
( 0,216)\(^4\) = \([\left(0,6\right)^3]^4\) = (0.6)\(^{3.4}\) = ( 0,6)\(^{12}\)
\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)
\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)
=> \(x=\left(\dfrac{3}{4}\right)^5\)
b, \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
\(x=\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\Leftrightarrow x=\left(\dfrac{2}{5}\right)^{8-5}\)
=>\(x=\left(\dfrac{2}{5}\right)^3\)
bài 2 : Với bài này ta cần áp dụng quy tắc: \(\left(x^m\right)^n=x^{m.n}\)
\(0,36^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)
\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)
a)\(\left(\dfrac{-3}{7}+\dfrac{5}{11}\right):\dfrac{-3}{5}+\left(\dfrac{-9}{7}+\dfrac{6}{11}\right):\dfrac{-3}{5}\)
=\(\dfrac{2}{77}\):\(\dfrac{-3}{5}+\left(\dfrac{-57}{77}\right):\dfrac{-3}{5}\)
=[\(\dfrac{2}{77}+\left(\dfrac{-57}{77}\right)\)]:\(\dfrac{-3}{5}\)
=\(\left(\dfrac{_{ }-5}{7}\right):\dfrac{-3}{5}\)
=\(\dfrac{25}{21}\)
b)\(\left(x-\dfrac{1}{4}\right)^3=27\)
⇔\(\left(x-\dfrac{1}{4^{ }}\right)^3=3^3\)
⇔\(x-\dfrac{1}{4}=3\)
⇔\(x=3+\dfrac{1}{4}\)
⇔\(x=\dfrac{13}{4}\)
Vậy \(x=\dfrac{13}{4}\)
7)
\(\left(0,36\right)^8=\left(0,6^2\right)^8=0,6^{16}.\)
8)
a) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left[\left(\frac{3}{5}\right)^2\right]^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)
\(\Rightarrow n=10\)
Vậy \(n=10.\)
b) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-0,25\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow\left(-0,25\right)^p=\left(0,25\right)^4\)
\(\Rightarrow p=4\)
Vậy \(p=4.\)
Chúc bạn học tốt!
\(\frac{8^{11}.3^{17}}{27^{10}.9^{15}}=\frac{8^{11}.3^{17}}{3^{30}.3^{30}}=\frac{8^{11}}{3^{13}.3^{30}}=\frac{8^{11}}{3^{43}}\)
\(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{[\left(5-1\right).5^3]^3}{5^{12}}=\frac{\left(4.5^3\right)^3}{5^{12}}=\frac{64.5^9}{5^{12}}=\frac{64}{5^3}=\left(\frac{4}{5}\right)^3\)
\(\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+6^{20}}{6^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{30}\right)}{3^{20}.\left(2^{20}-2+3^{20}\right)}=\frac{2^{20}}{3^{20}}=\left(\frac{2}{3}\right)^{20}\)
a) (0,25)^7 . 4^7 => (0,25.4)^7 = 1^7
b) 27^5 : 9^6 => (3^3)^5 : (3^2)^6 = 3^15 : 3^12 =3^3
c) (1/5)^11 . 5^10 => 5^-11 . 5^10 = 0.2 = 1/5 = 5^-1
chúc bạn hc tốt