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a) \(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2\)
=> \(\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}\)
=> \(\orbr{\begin{cases}3x=\frac{13}{22}\\3x=\frac{9}{22}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{13}{66}\\x=\frac{3}{22}\end{cases}}\)
b) \(\left(5-3x\right)^3=\left(-\frac{1}{27}\right)=\left(-\frac{1}{3}\right)^3\)
=> \(5-3x=-\frac{1}{3}\)
=> \(3x=\frac{16}{3}\)
=> \(x=\frac{16}{3}:3=\frac{16}{9}\)
c) 5x + 5x+2 = 650
=> 5x + 5x . 52 = 650
=> 5x(1 + 52) = 650
=> 5x . 26 = 650
=> 5x = 25
=> 5x = 52 => x = 2
d) 3x-1 + 5.3x-1 = 126
=> (1 + 5).3x-1 = 126
=> 6.3x-1 = 126
=> 3x-1 = 21
=> 3x-1 =3.7
tới đây là không xử lí được x luôn :)
a,\(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2=\left(-\frac{1}{11}\right)^2\)
\(< =>\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}< =>\orbr{\begin{cases}3x=\frac{1}{11}+\frac{1}{2}\\3x=-\frac{1}{11}+\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}3x=\frac{2}{22}+\frac{11}{22}=\frac{13}{22}\\3x=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{13}{22}:3=\frac{13}{22}.\frac{1}{3}=\frac{13}{66}\\x=\frac{9}{22}:3=\frac{9}{22}.\frac{1}{3}=\frac{9}{66}=\frac{3}{22}\end{cases}}\)
b,\(\left(5-3x\right)^2=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)
\(< =>5-3x=-\frac{1}{3}< =>-3x=-\frac{1}{3}-5=-\frac{16}{3}\)
\(< =>3x=\frac{16}{3}< =>x=\frac{16}{3}:3=\frac{16}{3}.\frac{1}{3}=\frac{16}{9}\)
c,\(5^x+5^{x+2}=650< =>5^x+5^x.25=650\)
\(< =>5^x\left(25+1\right)=5^x=\frac{650}{36}=25< =>x=2\)
bạn nào giúp câu d
5.3x = 5.34
=> x = 4
5.34 = 7.35 - 2.35
= 35 . (7 - 2)
= 5. 35
=> 3x = 35
=> x = 5
c) \(3^2+2^4-\left(6^8:6^6-6^2\right)< 5^x< 125\)
\(=9+16-\left(6^{8-6}-36\right)< 5^x< 5^3\)
\(=25-\left(6^2-36\right)< 5^x< 5^3\)
\(=25-\left(36-36\right)< 5^x< 5^3\)
\(=25-0< 5^x< 5^3\)
\(=25< 5^x< 5^3\)
\(=5^2< 5^x< 5^3\)
Vì \(5^2=25\) và \(5^3=125\) nên \(x\) không thể thỏa mãn đề bài
⇒ \(x\) không thỏa mãn đề bài
a) \(2.5^2.3^2+\left\{\left[2.5^3-\left(5x+4\right).5\right]:\left(2^2.3.5\right)\right\}=453\)
\(2.25.9+\left\{\left[2.125-\left(5x+4\right).5\right]:\left(4.3.5\right)\right\}=453\)
\(50.9+\left\{\left[250-\left(5x+4\right).5\right]:60\right\}=453\)
\(450+\left\{\left[250-\left(5x+4\right).5\right]:60\right\}=453\)
\(\left[250-\left(5x+4\right).5\right]:60=453-450\)
\(\left[250-\left(5x+4\right).5\right]:60=3\)
\(250-\left(5x+4\right).5=3.60\)
\(250-\left(5x+4\right).5=180\)
\(\left(5x+4\right).5=250-180\)
\(\left(5x+4\right).5=70\)
\(5x+4=70:5\)
\(5x+4=14\)
\(5x=14-4\)
\(5x=10\)
\(x=10:5\)
\(x=2\)
Vậy \(x=2\)
b) \(\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=-2\\x-\frac{1}{3}=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left(-2\right)+\frac{1}{3}\\x=2+\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{7}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{7};\frac{7}{3}\right\}\)
a. 5x3^x= 8x19683+7x19683
5x3^x=(8+7)x19683
5x3^x=15x19683
5x3^x= 295245
3^x=295245:5
=59049
3^x= 3^10
vậy x=10
d.x-32=0:45
x-32=0
x=0+32
x=32
a: \(\Leftrightarrow x^2=\dfrac{-5}{2}\cdot\dfrac{-10}{9}=\dfrac{50}{18}=\dfrac{25}{9}\)
=>x=5/3hoặc x=-5/3
c: \(\Leftrightarrow4\left(x-\dfrac{5}{8}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)
=>x-5/8=1/4
hay x=2/8+5/8=7/8
d: \(\Leftrightarrow\left|x-3\right|=\dfrac{2}{5}+\dfrac{3}{5}=1\)
=>x-3=1 hoặc x-3=-1
=>x=4 hoặc x=2
e: =>1-1/2x=-3
=>1/2x=4
hay x=8
c, \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\left(5-3\right)=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\cdot2=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}=5^{11}\)
\(\Leftrightarrow x+3=11\)
\(\Leftrightarrow x=8\)
Vậy x = 8
d, \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+2^{x+4}+2^{x+5}=480\)
\(\Leftrightarrow2^x\left(1+2+2^2+2^3+2^4+2^5\right)=480\)
\(\Leftrightarrow2^x\cdot63=480\)
\(\Leftrightarrow2^x=\frac{160}{21}\)
\(\Leftrightarrow x\approx2,93\)
bạn nào chơi ff ko?