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a) 52x+1=125
=>52x+1= 53
=>2x+1=3
=>2x = 2
=>x =1
c)73<x <=93
=>343<x<=729
=> x = { 344;345;...;729}
d) (2x+3)4=81
=>(2x+3)4=34
=>2x+3 =3
=>2x =0
=> x = 0
nhớ tk nha
a)52x+1=125
52x+1=53
⏩ 2x+1=3
2x=4
x=4:2
x=2
b) (2x)2(2x) 3 =25.25
(2x) 2 (2x) 3 =210
(2x) 5=1024
(2x) 5 =45
2x=4
x=4 :2=2
d)(2x +3)4=81
(2x +3)4=34
2x+3=3
2x=0
x=0
Câu c bạn tự giải nhé
Nhớ tk giùm mình đấy
\(\frac{9^{14}.25^5.8^7}{18^{12}.\left(25^2\right)^3.24^3}\)=\(\frac{9^{14}.25^5.8^7}{\left(9.2\right)^{12}.25^6.\left(8.3\right)^3}\)
=\(\frac{9^{14}.25^5.8^7}{9^{12}.2^{12}.25^6.8^3.3^3}\)=\(\frac{9^2.25^{-1}.8^4}{2^{12}.3^3}\)
=\(\frac{3^4.0,04.8^4}{8^4.3^3}\)=\(\frac{3.0,04}{1}\)
=3.0,04=0,12
k mình nhé
Ta có \(5^x=125\)
\(\Rightarrow5^x=5^3\Rightarrow x=3\)
Vậy x = 3
b,\(3^{2x}=81\)
\(\Rightarrow3^{2x}=3^4\Rightarrow2x=4\Rightarrow x=2\)
Vậy x = 2
c,\(5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}=5^2.3+5^2.2\Rightarrow5^{2x-3}=5^2.\left(2+3\right)\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\Rightarrow2x=6\Rightarrow x=3\)
Vậy x = 3
a) \(\left(\frac{-1}{2}\right)+\left|\frac{-7}{6}\right|+\left(\frac{-7}{8}\right)=\frac{-1}{2}+\frac{7}{6}+\left(\frac{-7}{8}\right)=\frac{-5}{24}\)
b) \(\frac{3}{2}.\left|2+x\right|=\frac{5}{4}\)
\(\Rightarrow\left|2+x\right|=\frac{5}{4}:\frac{3}{2}=\frac{5}{6}\)
\(\Rightarrow2+x=\pm\frac{5}{6}\)
\(\Rightarrow\orbr{\begin{cases}2+x=\frac{5}{6}\Rightarrow x=\frac{5}{6}-2=\frac{-7}{6}\\2+x=\frac{-5}{6}\Rightarrow x=\frac{-5}{6}-2=\frac{-17}{6}\end{cases}}\)
Vậy \(x=\left\{\frac{-7}{6};\frac{-17}{6}\right\}\)
c) Ta có: \(\frac{3,7}{x}=\frac{-5}{1,2}\Rightarrow3,7.1,2:\left(-5\right)=x\)
\(\Rightarrow x=-0,888\)
d) Tự làm
A=(\(\frac{-1}{2}\))+|\(\frac{-7}{6}\)|+(\(\frac{-7}{8}\))
=(\(\frac{-1}{2}\))+\(\frac{-7}{6}\)+(\(\frac{-7}{8}\))
=\(\frac{2}{3}\)+(\(\frac{-7}{8}\))
=\(\frac{-5}{24}\)
Vậy x=\(\frac{-5}{24}\)
B=\(\frac{3}{2}\)x|2+x|=\(\frac{5}{4}\)
=|2+x|=\(\frac{5}{4}\):\(\frac{3}{2}\)
=\(\orbr{\begin{cases}2+x=\frac{5}{6}\\2+x=\frac{-5}{6}\end{cases}}\)
=\(\orbr{\begin{cases}x=\frac{5}{6}-2\\x=\frac{-5}{6}-2\end{cases}}\)
=\(\orbr{\begin{cases}x=-1\frac{1}{6}\\x=-2\frac{5}{6}\end{cases}}\)
Vậy x\(\in\){\(-1\frac{1}{6}\);\(-2\frac{5}{6}\)}
\(x^3=x\)
\(\Rightarrow x^3-x=0\)
\(\Rightarrow x.\left(x^2-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
vậy ...
mấy câu khác tương tự
5.\(\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}}\)
vậy.....
câu 6 tương tự câu 5
\(5.\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\left(2x-4\right)^6=\left(2x-4\right)^4\)
\(\left(2x-4\right)^6-\left(2x-4\right)^4=0\)
\(\left(2x-4\right)^4[\left(2x-4\right)^2-1]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-4\right)^4=0\\\left(2x-4\right)^2=1\end{cases}}\)
\(\Rightarrow x=2\text{hoặc}x=\frac{5}{2}\text{hoặc}x=\frac{3}{2}\)
a)vi (2x-6)^7=(2x-6)^5
=>2x-6=0 hoac 2x-6=1
=>x=3hoac x=7/2
b)(7-2x)^9=(7-2x)^3
=>7-2x=0 hoac 7-2x=1
=>x=7/2 hoac x=3
a) \(\left(2x-6\right)^7=\left(2x-6\right)^5\)
\(\Leftrightarrow\left[2\left(x-3\right)\right]^7=\left[2\left(x-3\right)\right]^5\)
\(\Leftrightarrow2^7\left(x-3\right)^7=2^5\left(x-3\right)^5\)
\(\Leftrightarrow128\left(x-3\right)^7=32\left(x-3\right)^5\)
\(\Leftrightarrow4\left(x-3\right)^7=\left(x-3\right)^5\)
\(\Leftrightarrow4\left(x-3\right)^7-\left(x-3\right)^5=0\)
\(\Leftrightarrow\left(x-3\right)^5\left[4\left(x-3\right)-1\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\4\left(x-3\right)-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\x=\frac{7}{2}\\x=\frac{5}{2}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\x=\frac{7}{2}\\x=\frac{5}{2}\end{cases}}\)
vào đây : https://coccoc.com/search/math#query=(+2x-5)%5E9%3D(2x-5)%5E6
(2x- 15)5 = (2x - 15)3
=> 2x - 15 = 0 ; 2x - 15 = 1
2x - 15 = 0
2x = 15
x = 7,5
2x - 15 = 1
2x = 16
x = 8
\(f\left(x\right)g\left(x\right)=0\Leftrightarrow f\left(x\right)=0\)hoặc \(g\left(x\right)=0\)