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1. Tính giá trị biểu thức:
\(777:7+1331:11^3\)
\(=111+11^3:11^3\)
\(=111+1\)
\(=112\)
1.
\(777:7+1331:11^3=111+1331:1331=111+1=112\)
2.
\(2^{x+3}+2^x=36\\ 2^x\left(1+2^3\right)=36\\ 2^x\cdot9=36\\ 2^x=4\\ x=2\)
\(3^{x+4}+3^{x+2}=270\\ 3^{x+2}\cdot\left(3^2+1\right)=270\\ 3^{x+2}\cdot10=270\\ 3^{x+2}=27\\ x+2=3\\ x=1\)
\(\left(2x-5\right)^5=3^{10}\\ \left(2x-5\right)^5=3^{2\cdot5}\\ \left(2x-5\right)^5=\left(3^2\right)^5\\ \left(2x-5\right)^5=9^5\\ 2x-5=9\\ 2x=14\\ x=7\)
3.
\(31^{11}< 32^{11}=\left(2^5\right)^{11}=2^{55}\\ 17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}>2^{55}\\ \Rightarrow31^{11}< 17^{14}\)
\(5^{300}=5^{2\cdot150}=\left(5^2\right)^{150}=25^{150}\\ 3^{453}=3^{3\cdot151}=\left(3^3\right)^{151}=27^{151}\\ 25^{150}< 25^{151}< 27^{151}\\ \Leftrightarrow5^{300}< 3^{453}\)
Tìm x biết \(\left(x-5\right)^4=\left(x-5\right)^6\)
So Sánh: 2711và 818; 19920 và 200315; 399và 1121
a/
\(37^{1320}=\left(37^2\right)^{660}=1369^{660}\)
\(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\)
\(\Rightarrow1363^{660}>1331^{660}\Rightarrow37^{1320}>11^{1979}\)
b/
\(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
\(\Rightarrow27^{11}>81^8\)
d/
\(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 9^{21}< 11^{21}\)
e/ \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(\Rightarrow5^{36}>11^{24}\)
g/ \(21^{15}=3^{15}.7^{15}\)
\(27.49^8=3^3.\left(7^2\right)^8=3^3.7^{16}\)
\(\frac{21^{15}}{27.49^8}=\frac{3^{15}.7^{15}}{3^3.7^{16}}=\frac{3^{12}}{7}>1\Rightarrow21^{15}>27.49^8\)
f/ \(199^{20}=\left(199^4\right)^5\)
\(2003^{15}=\left(2003^3\right)^5\)
\(2003^5>1990^5\)
\(\frac{1990^5}{199^4}=\frac{199^5.10^5}{199^4}=199.10^5>1\)
\(\Rightarrow2003^5>1990^5>199^4\Rightarrow2003^{15}>199^{20}\)
Bài 1 \(a)5^{36}=(5^3)^{12}=125^{12}\)
\(11^{24}=(11^2)^{12}=121^{12}\)
Vì 125 > 121 nên \(5^{36}>11^{24}\)
\(b)21^{15}=(3\cdot7)^{15}=3^{15}\cdot7^{15}\)
\(27^5\cdot49^8=(3^3)^5\cdot(7^2)^8=3^{15}\cdot7^{16}\)
Vì 15 < 16 nên \(3^{15}\cdot7^{15}< 3^{15}\cdot7^{16}\)
hay : \(21^{15}< 27^5\cdot49^8\)
Bài 2 tự làm
Chúc bạn học tốt
Bài 1:
a) \(x^{10}=1^x\Rightarrow\orbr{\begin{cases}x=1\\x=10\end{cases}}\)
b) \(x^{10}=x\Rightarrow x=1\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5.\left(2x-15\right)^3=\left(2x-15\right)^3\)
\(\left(2x-15\right)^2=1\Rightarrow x=8\)
Bài 2:
\(a;2^{16}=2^{13}\cdot2^3=2^{13}\cdot8>7\cdot2^{13}\)
\(b;49^8\cdot27^5=7^{16}\cdot3^{15}=21^{15}\cdot7>21^5\)
C;Ta có:\(199^{20}< 200^{20}=2^{20}\cdot10^{40}=2^{15}\cdot10^{40}\cdot2^5\)
\(2003^{15}>2000^{15}=2^{15}\cdot10^{45}=2^{15}\cdot10^{40}\cdot10^5\)
Vì 25<105 nên 19920<200315
\(d;3^{39}< 3^{40}=9^{20}< 11^{20}< 11^{21}\)
1) \(2^{x+1}\cdot2^{2014}=2^{2015}\)\(\Leftrightarrow2^{2014x+2014}=2^{2015}\)\(\Leftrightarrow2014x+2014=2015\)\(\Leftrightarrow x=\frac{1}{2014}\)
2) \(7x-2x=\frac{6^{17}}{6^{15}}+\frac{44}{11}\)\(\Leftrightarrow5x=6^2+4=36+4=40\)\(\Leftrightarrow x=\frac{40}{5}=8\)
3) \(3^x=9\)\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)
4) \(7x-x=\frac{5^{21}}{5^{19}}+3\cdot2^2-7^0\)\(\Leftrightarrow6x=5^2+3\cdot4-1=25+12-1=36\)\(\Leftrightarrow x=6\)
5) \(4^x=64\)\(\Leftrightarrow4^x=4^3\)\(\Leftrightarrow x=3\)
6) \(9^{x-1}=9\)\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=0\)
7) \(\frac{2^x}{2^5}=1\)\(\Leftrightarrow2^{x-5}=2^0\)\(\Leftrightarrow x-5=0\)\(\Leftrightarrow x=5\)
8) \(\left(5x-9\right)^3=216\)\(\Leftrightarrow\left(5x-9\right)^3=6^3\)\(\Leftrightarrow5x-9=6\)\(\Leftrightarrow5x=15\)\(\Leftrightarrow x=3\)
9) \(5\cdot3^{7x-11}=135\)\(\Leftrightarrow5.3^{7x-11}=5.3^3\)\(\Leftrightarrow3^{7x-11}=3^3\)\(\Leftrightarrow7x-11=3\)\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
10) \(2.3^x=19\cdot3^8-81^2\)\(\Leftrightarrow2.3^x=19\cdot3^8-3^8=18.3^8=2.3^{11}\)\(\Leftrightarrow3^x=3^{11}\Leftrightarrow x=11\)
Đây là cách làm của mình. Bạn có thể chỉnh sửa tuỳ ý theo cách làm của bạn nhé ^^
Học tốt ^3^