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1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
a) \(4.\left(\frac{-1}{3}\right)^3-6.\left(\frac{-1}{2}\right)^2+\left(\frac{1}{2}\right)^0\)
= \(4.\left(\frac{-1}{27}\right)-6.\left(\frac{1}{4}\right)+1\)
= \(\frac{-4}{27}-\frac{3}{2}+1\)
= \(\frac{-8}{54}-\frac{81}{54}+\frac{54}{54}\)
= \(\frac{-35}{54}\)
b) \(\left(3^2\right)^2-\left[\left(-5\right)^2\right]^2+\left[\left(-2\right)^3\right]^2\)
= \(3^{2.2}-\left(-5\right)^{2.2}+\left(-2\right)^{3.2}\)
= \(3^4-\left(-5\right)^4+\left(-2\right)^6\)
= \(81-625+64\)
= \(-480\)
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
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 :
\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)
đang bận làm để thông cảm nha có j kiếm lại chất xám mình giải cho