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Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
Gửi tạm trước 2 câu !
\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)
Trả lời :
\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21
Bài 1:
a)
\((\frac{3}{5})^2-[\frac{1}{3}:3-\sqrt{16}.(\frac{1}{2})^2]-(10.12-2014)^0\)
\(=\frac{9}{25}-(\frac{1}{9}-1)-1\)
\(=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}\)
b)
\(|-\frac{100}{123}|:(\frac{3}{4}+\frac{7}{12})+\frac{23}{123}:(\frac{9}{5}-\frac{7}{15})\)
\(=\frac{100}{123}:\frac{4}{3}+\frac{23}{123}:\frac{4}{3}=(\frac{100}{123}+\frac{23}{123}):\frac{4}{3}=1:\frac{4}{3}=\frac{3}{4}\)
c)
\(\frac{(-5)^{32}.20^{43}}{(-8)^{29}.125^{25}}=\frac{5^{32}.(2^2.5)^{43}}{(-2)^{3.29}.(5^3)^{25}}=\frac{5^{32}.2^{86}.5^{43}}{-2^{87}.5^{75}}\)
\(=\frac{5^{32+43}.2^{86}}{-2^{87}.5^{75}}=\frac{5^{75}.2^{86}}{-2^{87}.5^{75}}=-\frac{1}{2}\)
Bài 2:
a)
\(\frac{2}{3}-(\frac{3}{4}-x)=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b)
\((\frac{1}{2}-x)^2=(-2)^2=2^2\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}-x=-2\\ \frac{1}{2}-x=2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=\frac{-3}{2}\end{matrix}\right.\)
c)
\(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
\(|3x+\frac{1}{2}|=\frac{2}{3}+1=\frac{5}{3}\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{1}{2}=\frac{5}{3}\\ 3x+\frac{1}{2}=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{7}{18}\\ x=\frac{-13}{18}\end{matrix}\right.\)
d)
\(3^{2x+1}=81=3^4\)
\(\Rightarrow 2x+1=4\Rightarrow x=\frac{3}{2}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
Câu a) số lớn lắm
b) \(3^{-3}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^x=3^3\)
=> \(3^x=3^3:\frac{1}{27}=3^3:\left(\frac{1}{3}\right)^3=3^3:\frac{1^3}{3^3}=3^3\cdot3^3=3^6\)
=> x = 6
b) \(\left(7x+2\right)^{-1}=3^{-2}\)
=> \(\frac{1}{7x+2}=\frac{1}{9}\)
=> 7x + 2 = 9
=> 7x = 7
=> x = 1
Bài 2:
a) \(3^4\cdot\frac{1}{729}\cdot81^3\cdot\frac{1}{9^2}\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot\left(3^4\right)^3\cdot\left(\frac{1}{3}\right)^4\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot3^{12}\cdot\left(\frac{1}{3}\right)^4=3^{16}\cdot\left(\frac{1}{3}\right)^{10}=\frac{3^{16}}{3^{10}}=3^6\)
b) \(\left(8\cdot2^5\right):\left(2^4\cdot\frac{1}{32}\right)=\left(2^3\cdot2^5\right):\left(2^4\cdot\left(\frac{1}{2}\right)^5\right)\)
\(=2^8:\left(2^4\cdot\frac{1^5}{2^5}\right)=2^8:\left(\frac{2^4}{2^5}\right)=2^8:2^{-1}=512\)
c) \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
d) Tương tự
Trả lời rõ cho mik có đc k mn