Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1;Ta có\(5.3^x=5.3^4\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
2.Ta có \(9.5^x=6.5^6+3.5^6\)
\(\Rightarrow9.5^x=5^6.\left(6+3\right)\)
\(\Rightarrow9.5^x=9.5^6\)
\(\Rightarrow5^x=5^6\)\
\(\Rightarrow x=6\)
3, Ta có \(2.3^{x+2}+4.3^{x+1}=10.3^6\)
\(\Rightarrow3^{x+1}.\left(2.3+4\right)=10.3^6\)
\(\Rightarrow3^{x+1}.10=10.3^6\)
\(\Rightarrow3^{x+1}=3^6\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
a) 5.3x = 5.34
=> 3x=34
=> x=4
b) 9.5x=6.56+3.56
=> 9.5x = (6+3)56
=> 9.5x=9.56
=> 5x=56
=> x=6
c) 2.3x+2 + 4.3x+1 = 10.36
=> 2.3x+1.3 + 4.3x+1 = 10.36
=> 6.3x+1+4.3x+1=10.36
=> (6+4).3x+1=10.36
=> 10.3x+1=10.36
=> 3x+1=36
=> x+1=6
=> x=5
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(5^{x+3}\left(5-3\right)=2.5^{11}\)
\(5^{x+3}.2=2.5^{11}\)
\(5^{x+3}=5^{11}\)
\(x+3=11\)
\(x=8\)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)
\(4^{x+1}.13=13.4^{11}\)
\(4^{x+1}=4^{11}\)
\(x+1=11\)
\(x=10\)
a)\(\frac{1}{4}-\frac{1}{3}x=\frac{2}{5}-\frac{3}{2}x\)
\(\Leftrightarrow\)\(\frac{15-20x}{60}=\frac{24-90x}{60}\)
\(\Leftrightarrow15-20x=24-90x\)
\(\Leftrightarrow-20x+90x=24-15\)
\(\Leftrightarrow70x=9\)
\(\Leftrightarrow x=\frac{9}{70}\)
c) (1/2-1/6)*3^x+4-4*3^x=3^16-4*3^13
=1/3*3^x*3^4-4*3^x=3^13*3^3-4*3^13
=27*3^x-4*3^x=3^13*(27-4)
=3^x*(27-4)=3^13*(27-4)
=>x=13
Câu a đề thiếu vế phải rồi bạn
b: \(\Leftrightarrow x\cdot0+1=0\)
=>0x+1=0(vô lý)
a) \(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b) \(x=\frac{5}{7}+\frac{2}{5}=\frac{39}{35}\)
c) \(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\Leftrightarrow x=\frac{4}{21}\)
d) \(x=\frac{4}{7}-\frac{1}{3}=\frac{5}{21}\)
a/ \(x+\frac{1}{3}=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b/\(x-\frac{2}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{2}{5}=\frac{39}{35}\)
c/\(-x-\frac{2}{3}=-\frac{6}{7}\)
\(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\)
\(\rightarrow x=\frac{4}{21}\)
d/ \(-\frac{4}{7}-x=\frac{1}{3}\)
\(x=\left(-\frac{4}{7}\right)-\frac{1}{3}=-\frac{19}{21}\)
#)Giải :
a)\(2^6.5^6=\left(2.5\right)^6=10^6\)
b)\(8^2.5^2=\left(8.5\right)^2=40^2\)
c)\(4^3.5^3=\left(4.5\right)^3=20^3\)
d)\(5^2.6^2.3^2=\left(5.6.2\right)^2=60^2\)
e)\(\frac{625^5}{25^8}=\frac{\left(25^2\right)^5}{25^8}=\frac{25^{10}}{25^8}=25^2\)
g)\(\frac{3^9}{7}.\frac{7^9}{3}=\frac{\left(3.7\right)^9}{7.3}=\frac{21^9}{21}=21^8\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
a) Ta có: \(3\cdot5^x=3\cdot5^4\)
\(\Leftrightarrow5^x=5^4\)
\(\Leftrightarrow x=4\)
Vậy: x=4
b) Ta có: \(9\cdot7^x=6\cdot7^5+3\cdot7^5\)
\(\Leftrightarrow9\cdot7^x=\left(6+3\right)\cdot7^5\)
\(\Leftrightarrow9\cdot7^x=9\cdot7^5\)
\(\Leftrightarrow7^x=7^5\)
\(\Leftrightarrow x=5\)
Vậy: x=5
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^7\)
\(\Leftrightarrow2\cdot3^{x+2}+2\cdot6^{x+1}=10\cdot3^7\)
\(\Leftrightarrow2\cdot\left(3^{x+2}+6^{x+1}\right)=10\cdot3^7\)
\(\Leftrightarrow3^{x+2}+6^{x+1}=\frac{10\cdot3^7}{2}\)
\(\Leftrightarrow3^{x+1}\cdot3+3^{x+1}\cdot2=5\cdot3^7\)
\(\Leftrightarrow3^{x+1}\cdot\left(3+2\right)=5\cdot3^7\)
\(\Leftrightarrow5\cdot3^{x+1}=5\cdot3^7\)
\(\Leftrightarrow3^{x+1}=3^7\)
\(\Leftrightarrow x+1=7\)
\(\Leftrightarrow x=6\)
Vậy: x=6
a) \(3.5^x=3.5^4\)
\(\Rightarrow3.5^x-3.5^4=0\)
\(\Rightarrow3.\left(5^x-5^4\right)=0\)
Vì \(3\ne0.\)
\(\Rightarrow5^x-5^4=0\)
\(\Rightarrow5^x=0+5^4\)
\(\Rightarrow5^x=5^4\)
\(\Rightarrow x=4\left(TM\right).\)
Vậy \(x=4.\)
Chúc bạn học tốt!