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\(4^{x+1}.2=32\)
\(4^{x+1}=32:2\)
\(4^{x+1}=16\)
\(4^{x+1}=4^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
vậy \(x=1\)
\(\left(x-\frac{2}{3}\right)^2=\frac{25}{81}\)
\(\left(x-\frac{2}{3}\right)^2=\left(\frac{5}{9}\right)^2\)
\(\Rightarrow x-\frac{2}{3}=\frac{5}{9}\)
\(\Rightarrow x=\frac{11}{9}\)
vậy \(x=\frac{11}{9}\)
\(500^{300}=\left(500^3\right)^{100}=125000000^{100}\)
\(300^{500}=\left(300^5\right)^{100}\)
vì \(\left(500^3\right)^{100}< \left(300^3\right)^{100}\)nên\(500^{300}< 300^{500}\)
\(4^{45}=\left(4^9\right)^5=262144^5\)
\(3^{60}=\left(3^{12}\right)^5=531441^5\)
vì \(262144^5< 531441^5\) nên \(4^{45}< 3^{60}\)
a) Thiếu đề (hoặc sai)
b) x đâu?
c)\(3x-1=x+2\)
\(\Rightarrow3x-x=2+1\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)
c) \(\frac{x+2}{5}=\frac{2-3x}{3}\)
\(\Rightarrow3.\left(x+2\right)=5.\left(2-3x\right)\)
\(\Rightarrow3x+6=10-15x\)
\(\Rightarrow3x+15x=10-6\)
\(\Rightarrow18x=4\)
\(\Rightarrow x=\frac{4}{18}=\frac{2}{9}\)
câu 1 là \(x\times\left(4.6+\frac{3}{5}\right)=7.2-8.15\)
câu 2 là \(42+\frac{3}{7}.\left[3\times x-1=12\right]\)
1. \(\left(\frac{1}{2}\right)^n=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^n=\frac{1^5}{2^5}\)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^5\)
Vậy \(n=5\)
2. \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
Vậy \(n=3\)
3. \(\frac{16}{2^n}=2\)
\(2^n=\frac{16}{2}\)
\(2^n=8=2^3\)
Vậy \(n=3\)
1. (1/2)2 = 1/32 <=> (21)n = (25)n <=> 1.n = 5.1 <=> n = 5
=> n = 5
2) 343/125 = (7/5)n <=> (7/5)3 = (7/5)n <=> 3 = n
=> n = 3
3) 16/2n = 2 <=> 16.2n <=> 2n = 2/16 <=> 2n = 1/8 <=> 2n = 8 <=> 2n = 23 <=> n = 3
=> n = 3
#)Giải :
Bài 1 :
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\Leftrightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Leftrightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2C=1-\frac{1}{3^{100}}\Leftrightarrow C=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\Rightarrow C< \frac{1}{2}\left(đpcm\right)\)
Bài 2 :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)+...+\left(\frac{1}{81}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)