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Bài 1 )
a ) \(2.2^2.2^3.....2^x=1024\Leftrightarrow2^{1+2+....+x}=2^{10}\Leftrightarrow1+2+....+x=10\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=10\Leftrightarrow\left(x+1\right)x=20=4.5\Rightarrow x=4\)
b ) \(\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow3x+39=259-7x\Leftrightarrow3x+7x=259-39\Leftrightarrow10x=220\Rightarrow x=22\)
Bài 2 ) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(\frac{50^2-15.125}{5^4}\right)^{2014}=\frac{1}{2}.8-\frac{2}{5}+\left(\frac{5^4.2^2-3.5^4}{5^4}\right)^{2014}\)
\(=4-\frac{2}{5}+\left[\frac{5^4\left(4-3\right)}{5^4}\right]^{2014}=\frac{18}{5}+1=\frac{23}{5}\)
Mình làm bài 1 thui nha, còn bài 2 thì còn tự tính là được thôi mừ !!!
Bài 1:
a) \(2.2^2.2^3...2^x=1024\)
\(=>2^{1+2+3+...+x}=2^{10}\)
\(< =>1+2+3+...+x=10\)
\(=>6+x=10\)
\(=>x=10-6\)
\(=>x=4.\)
Nếu đúng thì k cho mình nhá
a)\(\left(\frac{2}{5}\right)^{2014}:\left(\frac{4}{25}\right)^{1007}=\left[\left(\frac{2}{5}\right)^2\right]^{1007}:\left(\frac{4}{25}\right)^{1007}\)
\(=\left(\frac{4}{25}\right)^{1007}:\left(\frac{4}{25}\right)^{1007}\)
\(=1\)
b)\(3^{n+1}:9=3^{n+1}:3^2\)
\(=3^{n+1-2}\)
\(=3^{n-1}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
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
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =