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4,
a,\(\dfrac{x-1}{9}\)=\(\dfrac{8}{3}\)
[x- 1].3=9.8
[x- 1].3=72
x-1=72:3
x-1=24
x=24+1
x=25
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^{x+1}+3^x=108\)
\(3^x\cdot3+3^x=108\)
\(3^x\cdot\left(3+1\right)=108\)
\(3^x=27=3^3\)
\(\Rightarrow x=3\)
\(3^{x+1}+3^x=108\)
(=)\(3^x.3+3^x=108\)
(=)\(3^x.\left(3+1\right)=108\)
(=)\(3^x.4=108\)
(=)\(3^x=27\)
(=)\(3^x=3^3\)
=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
e) \(\frac{-84}{14}< 3x< \frac{108}{9}\)
\(\Rightarrow-2< x< 4\)
Vậy \(-2< x< 4\)
d) \(\frac{x-3}{y-3}=\frac{4}{3}\)
Ta có: x=y+5
\(\Rightarrow\frac{y+5-3}{y-3}=\frac{4}{3}\)
\(\Rightarrow\frac{y+2}{y-3}=\frac{4}{3}\)
\(\Rightarrow\left(y+2\right).3=\left(y-3\right).4\)
\(3y+6=4y-12\)
\(6+12=4y-3y\)
\(y=18\)
\(\Rightarrow x=5+y=5+18=23\)
KL:............................
a) \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\)\(\left(x-1\right).3=9.8\)
\(3x-3=72\)
\(3x=75\)
\(x=\frac{75}{3}\)
\(x=25\)
Câu b tương tự
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x.\left(x+1\right)=4.18\)
\(x.\left(x+1\right)=72\)
\(x.\left(x+1\right)=8.9\)
\(\Rightarrow x=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\frac{27^4.8^{17}}{9^6.32^3}=\frac{\left(3^3\right)^4.\left(2^3\right)^{17}}{\left(3^2\right)^6.\left(2^5\right)^3}=\frac{3^{12}.2^{51}}{3^{12}.2^{15}}=\frac{3^{12}.2^{15}.2^{36}}{3^{12}.2^{15}}=2^{36}\)
\(B=\frac{72^3.54^3:8^3}{108^5:4^5}=\frac{\left(72.54:8\right)^3}{\left(108:4\right)^5}=\frac{486^3}{27^5}=\frac{\left(3^5.2\right)^3}{\left(3^3\right)^5}=\frac{3^{15}.2^3}{3^{15}}=2^3=8\)
Bài 2
A = 2 +22 + 23 + 24 + ....+ 2100
A = ( 2+22 ) + (23 + 24 ) + ....+ (299 + 2100 )
A = 2(1+2 ) + 23 (1+2 ) + ...+ 299(1+2)
A = 2.3 + 23.3 + ....+ 299 .3
A = 3(2+23 + ...+ 299 )
=> A \(⋮\) 3 ( đpcm )
Bài 3
a, 2.3x = 312 .34 + 20 .274
2.3x = 312 . 34 + 20 . (33 ) 4
2.3x = 312 .34 + 20 .312
2.3x = 312(34+20 )
2.3x = 312 . 54
2.3x = 312 . 27 .2
2.3x = 312 . 33 .2
2.3x = 315 .2
=> x=15
b , (2x +1 ) 2 + 3.(22 + 1 ) = 22 .10
(2x +1 ) 2 + 3.(4+1 ) = 4.10
(2x +1 ) 2 + 3.5 = 40
(2x +1 ) 2 + 15 = 40
(2x +1 ) 2 = 40-15
(2x +1 ) 2 = 25
(2x +1 ) 2 = 52
=> 2x + 1 = 5
2x = 5-1
2x = 4
2x = 22
=> x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)
= \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)
= \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)
= \(\frac{2^8.\left(-6\right)}{2^8}\)
= \(-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x :
a) \(2.x.\frac{-3}{4}=-\frac{5}{12}\)
\(\Rightarrow2x=-\frac{5}{12}:-\frac{3}{4}\)
\(\Rightarrow2x=\frac{5}{9}\)
\(\Rightarrow x=\frac{5}{9}:2\)
\(\Rightarrow x=\frac{5}{18}\)
Vậy : \(x=\frac{5}{18}\)
b) \(\frac{2}{3}+\frac{1}{3}.x=7\)
\(\Rightarrow\frac{1}{3}.x=7-\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}.x=\frac{19}{3}\)
\(\Rightarrow x=\frac{19}{3}:\frac{1}{3}\)
\(\Rightarrow x=19\)
Vậy : \(x=19\)
c) \(\left(4.x+\frac{1}{8}\right)=\frac{3}{10}\)
\(\Rightarrow4.x=\frac{3}{10}-\frac{1}{8}\)
\(\Rightarrow4.x=\frac{7}{40}\)
\(\Rightarrow x=\frac{7}{40}:4\)
\(\Rightarrow x=\frac{7}{160}\)
Vậy : \(x=\frac{7}{160}\)
d) \(\frac{1}{3}.x-5=1\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}.x-5=\frac{3}{2}\)
\(\Rightarrow\frac{1}{3}.x=\frac{3}{2}+5\)
\(\Rightarrow\frac{1}{3}.x=\frac{13}{2}\)
\(\Rightarrow x=\frac{13}{2}:\frac{1}{3}\)
\(\Rightarrow x=\frac{39}{2}\)
Vậy : \(x=\frac{39}{2}\)
e) \(-\frac{2}{3}.x+\frac{1}{3}=-\frac{1}{2}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{1}{2}-\frac{1}{3}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{5}{6}\)
\(\Rightarrow x=-\frac{5}{6}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=\frac{5}{4}\)
Vậy : \(x=\frac{5}{4}\)
\(3^{x+1}+3^x=108\)
\(\Rightarrow3^x\left(1+3\right)=108\)
\(\Rightarrow3^x.4=108\)
\(\Rightarrow3^x=108:4\)
\(\Rightarrow3^x=27\)
\(27=3^3\)
\(3^x=3^3\)
\(\Rightarrow x=3\)
Vậy x = 3
3\(^{x+1}\)+ 3\(^x\)=108
3\(^x\).3 + 3\(^x\)= 108
3\(^x\)( 3 + 1 ) = 108
3\(^x\). 4 = 108
3\(^x\)= 27 = 3\(^3\)
x = 3