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![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^x-2-3^2=2^4-\left(2^8x2^4-2^{10}x2^2\right)\)
\(5^x-2-3^2=2^4-\left(2^{8+4}-2^{10+2}\right)\)
\(5^x-2-3^2=2^4-\left(2^{12}-2^{12}\right)\)
\(5^x-2-3^2=2^4-0\)
\(5^x-2-3^2=2^4\)
\(5^x-2-9=16\)
\(5^x-2=16+9\)
\(5^x-2=25\)
\(5^x=25+2\)
\(5^x=27\)
Bởi vì 27 không phân tích được 1 số có số mũ là 2
\(\Rightarrow\) Không tồn tại x
\(5^{x-2}-9=16-\left(256.16-1024.4\right)\)
\(\Rightarrow5^{x-2}-9=16-\left(4096-4096\right)\)
\(\Rightarrow5^{x-2}-9=16-0\)
\(\Rightarrow5^{x-2}-9=16\)
\(\Rightarrow5^{x-2}=25\)
\(\Rightarrow x-2=25:5\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 10-x-5=-5-7-11
=> 5 - x = -23
=> x = 28
b) |x| -3=0
=> |x| = 3
=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0
=> 7 - |x| = 0 hoặc 2x - 4 = 0
=> |x| = 7 hoặc 2x = 4
=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19
=> 2 + 3x = -34
=> 3x = 36
=> x = 12
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Rightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy x = 2
b) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Rightarrow-3x=-48\)
\(\Leftrightarrow x=16\)
Vậy x = 16
c) \(\dfrac{1}{9}=\dfrac{-2x}{10}\)
\(\Rightarrow-18x=10\)
\(\Leftrightarrow x=-\dfrac{5}{9}\)
Vậy \(x=-\dfrac{5}{9}\)
d) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{x}-5=\dfrac{-9}{x}+2\)
\(\Leftrightarrow\dfrac{3-5x}{x}=\dfrac{-9+2x}{x}\)
\(\Rightarrow3-5x=-9+2x\)
\(\Leftrightarrow7x=12\)
\(\Leftrightarrow x=\dfrac{12}{7}\)
Vậy \(x=\dfrac{12}{7}\)
e) ĐKXĐ: \(x\ne0\)
\(\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Rightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy \(x=\pm4\)
a) Ta có: \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Leftrightarrow-x=\dfrac{6\cdot\left(-8\right)}{3}=-16\)
hay x=16
Vậy: x=16
![](https://rs.olm.vn/images/avt/0.png?1311)
A. x = 2
B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)
C. x = 3
D. \(x=\dfrac{4.6}{8}=3\)
E. \(x=\dfrac{7}{3}\)
G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)
<=>\(14\left(10-x\right)=364\)
<=> 10 - x = 26
<=> x = -16
H. \(3\left(x+2\right)=4\left(x-5\right)\)
<=> 3x + 6 = 4x - 20
<=> -x = -26
<=> x = 26
K. \(\dfrac{x}{2}=\dfrac{8}{x}\)
<=> \(x^2=16\)
<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
M. \(\left(x-2\right)^2=100\)
<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
a=2
b=16
c=3
d=3
mik chỉ biết thế này thôi(ko chắc đúng=3)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{7}< \dfrac{x}{35}< \dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{35}< \dfrac{x}{35}< \dfrac{14}{35}\)
\(\Rightarrow5< x< 14\)
b) \(\dfrac{5}{13}< 2-x< \dfrac{5}{8}\)
\(\Rightarrow2-\dfrac{5}{8}< x< 2-\dfrac{5}{13}\)
\(\Rightarrow\dfrac{11}{8}< x< \dfrac{21}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
cau b ta co9;(x+x+.....+x)-(1+2+....+100)=4950
100x - 5050 =4950
100x =4950+5050
100x =10000
x=10000 ;100
x=100
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2^4:8+5.3^{x+1}\right).2=94\)
\(\Rightarrow\left(2+5.3^{x+1}\right)=94:2=47\)
\(\Rightarrow5.3^{x+1}=47-2=45\)
\(\Rightarrow3^{x+1}=9=3^2\Rightarrow x=2-1=1\)
(24:8 + 5.3x+1) . 2 = 94
2 + 5.3x.3 = 47
15.3x = 45
3x = 3 = 31
=> x = 1