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a. 2\(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}:2^0\right):2^3\)
= \(\left(\frac12+\frac13\right):\left(\frac12-\frac13\right)+\left(\frac12:1\right):8\)
=\(\left(\frac36+\frac26\right):\left(\frac36-\frac26\right)+\left(\frac12\right):8\)
= \(\left(\frac56:\frac16\right)+\left(\frac{1}{2_{}}\right):8\)
= \(5+\frac12:8\)
= \(5+\frac{1}{16}\)
= \(\frac51+\frac{1}{16}=\frac{80}{16}+\frac{1}{16}=\frac{81}{16}\)
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
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c: \(=\dfrac{7}{23}\cdot\dfrac{-24-45}{18}=\dfrac{7}{23}\cdot\dfrac{-69}{18}=\dfrac{7}{18}\cdot\left(-3\right)=-\dfrac{7}{6}\)
d: \(=\dfrac{7}{5}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{7}{5}\cdot10=14\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{9^5}=3^{40}-1\)
đăng từng câu nhé bạn
chứ kiểu vậy thì ko có ai giải cho bạn đâu
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
A = 1/2 + 1/4 + 1/8 + ... + 1/128
A = 1/2^1 + 1/2^2 + 1/2^3 + ... + 1/2^7
2A = 1 + 1/2 + 1/2^2 + ... + 1/2^6
2A - A = 1 - 1/2^7 = A
G = 5 - 5^2/1*6 5^2/6*11 - ... - 5^2/101*106
G = -5(-1 + 5/1*6 + 5/6*11 + ... + 5/101*106)
G = -5(-1 + 1 - 1/6 + 1/6 - 1/11 + ... + 1/101 - 1/106)
G = -1.(-1/106)
G = 1/106
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A-A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
D = 6/1*3 + 6/3*5 + 6/5*7 + ... + 6/2009*2011
D = 3(2/1*3 + 2/3*5 + 2/5*7 + ... + 2/2009*2011)
D = 3(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2009 - 1/2011)
D = 3(1 - 1/2011)
D = 3*2010/2011 = 6030/2011
F = (1 + 1/2)(1 + 1/3)(1 + 1/4)...(1 + 1/100)
F = 3/2*4/3*5/4*...*101/100
F = 101/2
Vậy_
\(A=\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{64}\)
\(\Rightarrow A=1-\frac{1}{128}=\frac{127}{128}\)
\(\frac{2^2}{3}\cdot\frac{3^2}{2\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{59^2}{58\cdot60}\)
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{59\cdot59}{58\cdot60}\)
\(=\frac{\left(2\cdot3\cdot\cdot\cdot\cdot\cdot59\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot59\right)}{\left(1\cdot2\cdot\cdot\cdot\cdot\cdot58\right)\cdot\left(3\cdot4\cdot\cdot\cdot\cdot\cdot60\right)}\)
\(=\frac{59\cdot2}{1\cdot60}=\frac{59}{30}\)
\(D=\frac{6}{1\cdot3}+\cdot\cdot\cdot+\frac{6}{2009\cdot2011}\)
\(=3\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdot\cdot\cdot+\frac{2}{2009\cdot2011}\right)\)
\(=3\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=3\cdot\left(1-\frac{1}{2011}\right)\)
\(=3\cdot\frac{2010}{2011}=\frac{6030}{2011}\)
\(\frac{1\cdot2\cdot3+2\cdot4\cdot6-3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot7+2\cdot6\cdot14-3\cdot9\cdot21+5\cdot15\cdot35}\)
\(=\frac{2\cdot\left(1\cdot3+4\cdot6-3\cdot3\cdot9+5\cdot5\cdot15\right)}{7\cdot\left(1\cdot3+4\cdot6-3\cdot3\cdot9+5\cdot5\cdot15\right)}\)
\(=\frac{2}{7}\)
\(\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\cdot\cdot\cdot\cdot\left(1+\frac{1}{100}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\cdot\frac{101}{100}\)
\(=\frac{3\cdot4\cdot\cdot\cdot\cdot\cdot101}{2\cdot3\cdot\cdot\cdot\cdot\cdot100}\)
\(=\frac{101}{2}\)
\(5-\frac{5^2}{1\cdot6}-\cdot\cdot\cdot-\frac{5^2}{101\cdot106}\)
\(=5\cdot\left(1-\frac{5}{1\cdot6}-\cdot\cdot\cdot-\frac{5}{101\cdot106}\right)\)
\(=-5\cdot\left(-1+1-\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{101}-\frac{1}{106}\right)\)
\(=-5\cdot\left(\frac{-1}{106}\right)\)
\(=\frac{5}{106}\)