soanhs các số sau
a) 3100 và 2150
b) 275 và 2453
c) 8175 và 30100
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e) 3n+2 + 5.3n + 1 = 216
=> 3n . 32 + 5.3n . 31 = 216
=> 3n . 9 + 15.3n = 216
=> 3n ( 9 + 15) = 216
=> 3n . 24 = 216
=> 3n = 9
=> n = 2
f) 5n + 1 - 5n - 1 = 1254 . 23 . 37
=> 5n . 5 - 5n . 1/5 = 1254 . 23 . 37
=> 5n ( 5 - 1/5) = 1254 . 23 . 37
=> 5n . 24/5 = 1254 . 23 . 37
=> n không thỏa mãn
c) 3n + 2 - 3n + 1 = 1458
=> 3n . 32 - 3n . 31 = 1458
=> 3n (32 - 3) = 1458
=> 3n . 6 = 1458
=> 3n = 243
=> n = 5
d) 2n - 1 + 4.2n = 9.25
=> 2n . 21 + 4.2n = 9.25
=> 2n (2 + 4) = 9.25
=> 2n . 6 = 9.25
=> 2n = \(\frac{9\cdot2^5}{6}=48\)
=> không tìm được x
a.
\(5^n+5^{n+2}=650\)
\(5^n\left(1+5^2\right)=650\)
\(5^n\left(1+25\right)=650\)
\(5^n\cdot26=650\)
\(5^n=650:26\)
\(5^n=25\)
\(5^n=5^2\Rightarrow n=2\)
b.
\(3^{n+3}+5\cdot3^n=864\)
\(3^n\left(3^3+5\right)=864\)
\(3^n\left(27+5\right)=864\)
\(3^n\cdot32=864\)
\(3^n=864:32\)
\(3^n=27\)
\(3^n=3^3\Rightarrow n=3\)
a) 5n + 5n+2 = 650
=> 5n + 5n . 52 = 650
=> 5n (1 + 52) = 650
=> 5n . 26 = 650
=> 5n = 25
=> n = 2
b) 3n+ 3 + 5.3n = 864
=> 3n . 33 + 5.3n = 864
=> 3n(33 + 5) = 864
=> 3n . 32 = 864
=> 3n = 27
=> n = 3
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
a/
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(A=2A-A=1-\frac{1}{2^{100}}< 1\)
b/
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)
\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)
a.
\(\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{12}\right)+\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\left(\frac{5}{4}+\frac{3}{5}+\frac{3}{8}-\frac{7}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\frac{33}{40}:\left(-\frac{11}{12}\right)\)
\(=\frac{33}{40}\cdot\left(-\frac{12}{11}\right)\)
\(=\frac{-9}{10}\)
b.
\(\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{15}\right)+\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\left(\frac{3}{8}-\frac{7}{5}+\frac{5}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\frac{33}{40}:\left(-\frac{11}{15}\right)\)
\(=\frac{33}{40}\cdot\left(-\frac{15}{11}\right)\)
\(=\frac{-9}{8}\)
a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)
=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)
=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)
b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)
=> \(-\frac{7}{8}x-\frac{23}{4}=3\)
=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)
=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)
c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)
=> \(2x-\frac{2}{7}-7=-11\)
=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)
=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)
d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)
=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)
=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)
\(\frac{x+5}{x+3}< 1\)
=> \(\frac{x+3+2}{x+3}< 1\)
=> \(1+\frac{2}{x+3}< 1\)
=> \(\frac{2}{x+3}< 0\)
=> x + 3 < 0
=> x < -3
Vậy x < -3 để \(\frac{x+5}{x+3}< 1\)
x + y = 3y
=> x= 3y-y
\(\frac{1}{x}\)= \(\frac{1}{3y-y}\)
\(\frac{1}{x}\)+\(\frac{1}{y}\)=\(\frac{1}{3y-y}\)+\(\frac{1}{y}\)
= \(\frac{y}{y\left(3y-y\right)}+\frac{3y-y}{y\left(3y-y\right)}\)=\(\frac{y+3y-y}{3y^2-y^2}\)=\(\frac{3y}{y^2\left(3-1\right)}=\frac{3}{2y}\)
Ta có x+y=3y
\(\Rightarrow x=2y\)
Thay vào ta có
\(\frac{1}{2y}+\frac{1}{y}=\frac{1}{2y}+\frac{2}{2y}=\frac{3}{2y}\)
a)Ta có : \(\hept{\begin{cases}3^{100}=\left(3^2\right)^{50}=9^{50}\left(1\right)\\2^{150}=\left(2^3\right)^{50}=8^{50}\left(2\right)\end{cases}}\)
Mà 9 > 8 => 950 > 850 => 3100 > 2150
Vậy 3100 > 2150
b) Ta có : \(\hept{\begin{cases}27^5=\left(3^3\right)^5=3^{15}\left(3\right)\\243^3=\left(3^5\right)^3=3^{15}\left(4\right)\end{cases}}\)
Từ (3) và (4) => 315 = 315 hay 275 = 2433
Vậy 275 = 2433 ( nên sửa lại 245 --> 243 nhá)
c) Ta có : \(81^{75}=\left(3^4\right)^{75}=3^{300}=\left(3^3\right)^{100}=27^{100}\)
Mà 27 < 30 => 27100 < 30100 hay 8175 < 30100
Vậy 8175 < 30100
a.
\(3^{100}=\left(3^2\right)^{50}=9^{50}\)
\(2^{150}=\left(2^3\right)^{50}=8^{50}\)
\(9^{50}>8^{50}\)
\(\Rightarrow3^{100}>2^{150}\)
b.
\(27^5=\left(3^3\right)^5=3^{15}\)
\(243^3=\left(3^5\right)^3=3^{15}\)
\(3^{15}=3^{15}\)
\(\Rightarrow27^5=243^3\)
c.
\(81^{75}=\left(3^4\right)^{75}=3^{300}=\left(3^3\right)^{100}=27^{100}\)
\(27^{100}< 30^{100}\Rightarrow81^{75}< 30^{100}\)