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29 tháng 9 2016
a) \(3^{x+1}-2.3^x=81\)
\(\Rightarrow3^x.3-2.3^x\)
\(\Rightarrow3^x\left(3-2\right)=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
b) \(2^{x+2}+2^x=40\)
\(\Rightarrow2^x.2^2+2^x=40\)
\(\Rightarrow2^x.\left(2^2+1\right)=40\)
\(\Rightarrow2^x.5=40\)
\(\Rightarrow2^x=8\)
\(\Rightarrow2^x=2^3\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
c) \(\left(x+1\right)^5-12=20\)
\(\Rightarrow\left(x+1\right)^5=32\)
\(\Rightarrow\left(x+1\right)^5=2^5\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
d) \(56-\left(x-2\right)^2=20\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow x-2=\pm6\)
+) \(x-2=6\Rightarrow x=8\)
+) \(x-2=-6\Rightarrow x=-4\)
Vậy \(x=8\) hoặc \(x=-4\)
23 tháng 9 2016
\(A=1+3+3^2+...+3^{2016}\)
\(3A=3.\left(1+3+3^2+...+3^{2016}\right)\)
\(3A=3+3^2+3^3+...+3^{2017}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2017}\right)-\left(1+3+3^2+...+3^{2016}\right)\)
\(2A=3^{2017}-1\)
\(A=\left(3^{2017}-1\right):2\)
\(B=1+6+6^2+...+6^{200}\)
\(6B=6.\left(1+6+6^2+...+6^{200}\right)\)
\(6B=6+6^2+6^3+...+6^{201}\)
\(6B-B=\left(6+6^2+6^3+...+3^{201}\right)-\left(1+6+6^2+...+6^{200}\right)\)
\(5B=6^{201}-1\)
\(B=\left(6^{201}-1\right):5\)
23 tháng 9 2016
\(3^{x-2}.4=324\)
\(3^{x-2}=324:4\)
\(3^{x-2}=81\)
\(3^{x-2}=3^4\)
\(x-2=4\)
\(x=4+2\)
\(x=6\)
\(2x< 20\)
\(\Rightarrow x=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
8 tháng 5 2018
\(a)\frac{2}{3}x-\frac{1}{2}=\frac{5}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{5}{12}+\frac{1}{2}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{2}{3}=\frac{11}{8}\)
\(b)\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Rightarrow\frac{14}{5}x-50=51.\frac{2}{3}=34\)
\(\Rightarrow\frac{14}{5}x=34+50=84\)
\(\Rightarrow x=84:\frac{14}{5}=30\)
8 tháng 5 2018
a) 2/3.x - 1/2 = 5/12
2/3.x = 5/12 + 1/2
2/3.x = 11/12
x = 11/12 : 2/3
x = 11/8
b) \(\left(2\frac{4}{5}.x-50\right):\frac{2}{3}=51\)
\(\frac{14}{5}.x-50=51.\frac{2}{3}\)
\(\frac{14}{5}.x-50=34\)
\(\frac{14}{5}.x=34+50\)
\(\frac{14}{5}.x=84\)
\(x=84:\frac{14}{5}\)
\(x=30\)
T
2 tháng 4 2018
Trước hết ta hãy so sánh :
\(\dfrac{10^{100}+1}{10^{101}+1}\)với \(\dfrac{10^{100}+1}{10^{102}+1}\)
Ta có: Cả hai phân số trên cùng tử.
\(\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{102}+1}\)
Tiếp đó so sánh : \(\dfrac{10^{101}+1}{10^{102}+1}\)với \(1\)
Ta được: \(\dfrac{10^{101}+1}{10^{102}+1}< 1\)
Ta lại so sánh được:\(\dfrac{10^{100}+1}{10^{102}+1}< 1\) (*)
Từ (*) suy ra \(\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+2}< \dfrac{10^{101}+1}{10^{102}+1}< 1\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+1}\)
Ngoài ra còn một cách như sau:
\(\dfrac{10^{101}+1}{10^{102}+1}=\dfrac{10^{\left(100+1\right)}+1}{10^{\left(101+1\right)}+1}=\dfrac{10}{10}.\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{101}+1}\) hay B > A hay A < B
ND
3 tháng 4 2018
Bài 1:
d)
\(\dfrac{x+5}{95}+\dfrac{x+10}{90}+\dfrac{x+15}{85}+\dfrac{x+20}{80}=-4\)
\(\Leftrightarrow\dfrac{x+5}{95}+1+\dfrac{x+10}{90}+1+\dfrac{x+15}{85}+1+\dfrac{x+20}{80}+1=-4+1+1+1+1\)
\(\Leftrightarrow\dfrac{x+100}{95}+\dfrac{x+100}{90}+\dfrac{x+100}{85}+\dfrac{x+100}{80}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\right)=0\)
\(\Leftrightarrow x+100=0\) ( vì: \(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\ne0\))
\(\Leftrightarrow x=-100\)
2 tháng 2 2020
Đề: X=\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+.......+\(\frac{1}{1+2+3+4+20}\)
X=\(\frac{1}{2.3:2}\)+\(\frac{1}{3.4:2}\)+\(\frac{1}{4.5:2}\)+......+\(\frac{1}{20.21:2}\)
X=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)\(\frac{2}{4.5}\)+........+\(\frac{2}{20.21}\)
X=2.(\(\frac{1}{2}\).3+\(\frac{1}{3}\).4+\(\frac{1}{4}\).5+.....+\(\frac{1}{20}\).21)
X=2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+......+\(\frac{1}{20}\)-\(\frac{1}{21}\))
X=2.(\(\frac{1}{2}\)-\(\frac{1}{21}\))
X=2.(\(\frac{21}{42}\)-\(\frac{2}{42}\))
X=2.\(\frac{19}{42}\)
X=\(\frac{19}{21}\)
Mn xem thử đúng ko nha!
CC
3 tháng 2 2020
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .......... ; \(1+2+3+....+20=\frac{20.21}{2}\)
\(\Rightarrow X=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.......+\frac{1}{\frac{20.21}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+........+\frac{2}{20.21}=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{20.21}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{20}-\frac{1}{21}\right)=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)
$20-2\left(x-1\right)^2=2$
=>\(2\left(x-1\right)^2=20-2=18\)
=>\(\left(x-1\right)^2=9\)
=>\(\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3+1=4\\x=-3+1=-2\end{matrix}\right.\)
20 - 2(x - 1)2 =2
2(x - 1)2= 20 - 2
2(x-1)2= 18
(x - 1)2 = 18 : 2
(x - 1)2 = 9
(x - 1 )2 = 32
x - 1 = 3
x = 3 + 1
x = 4
Vậy x= 4