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a, 42x - 6 = 1
=> 42 x = 7
=> x = 6
b, 5x + 5x + 1 +5x + 2 = 775
=> 15 x + 3 = 775
=> 15 x = 772
=> x = 772/ 15
5\(^{x+1}\) - 5\(^x\) = 2.28 + 8
5\(^x\).(5 - 1) = 520
5\(^x\).4 = 520
5\(^x\) = 520 : 4
5\(^x\) = 130
Với \(x\) = 0 ⇒ 5\(^x\) = 50 = 1 < 130 (loại)
Với \(x\) > 0 ⇒ 5\(^x\) = \(\overline{...5}\) \(\ne\) 130 (loại)
Vậy \(x\) \(\in\) \(\varnothing\)
\(5^{x+1}-5^x=2.2^8+8\\ 5^x\left(5-1\right)=512+8\\ 5^x.4=520\\ 5^x=\dfrac{520}{4}=130\)
Em xem lại đề
\(5^{x-2}+5^{x+2}=3130\\ 5^{x-2}.\left(1+5^4\right)=3130\\ 5^{x-2}.626=3130\\ 5^{x-2}=\dfrac{3130}{626}=5\\ Vậy:5^{x-2}=5\\ Vậy:x-2=1\\ Vậy:x=3\)
\(5^{x-2}+5^{x+2}=3130\)
\(\Rightarrow5^x\cdot\left(5^{-2}+5^2\right)=3130\)
\(\Rightarrow5^x\cdot\left(\dfrac{1}{25}+25\right)\)
\(\Rightarrow5^x\cdot\dfrac{626}{25}=3130\)
\(\Rightarrow5^x=3130:\dfrac{626}{25}\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy: x=3
\(a,12\left(x-1\right)=0\\ x-1=0\\ x=1\\ b,45+5\left(x-3\right)=70\\ 5\left(x-3\right)=25\\ x-3=5\\ x=8\\ c,3.x-18:2=12\\ 3.x-9=12\\ 3.x=21\\ x=7\)
\(\dfrac{5x}{1.6}+\dfrac{5x}{6.11}+\dfrac{5x}{11.16}+\dfrac{5x}{16.21}+\dfrac{5x}{21.26}+\dfrac{5x}{26.31}=1\)
\(=x\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\right)=1\)
\(=x\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\right)=1\)
\(=x\left(1-\dfrac{1}{31}\right)=1\)
\(\Rightarrow x=1:\left(1-\dfrac{1}{31}\right)=\dfrac{31}{30}\)
Ta có :
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)
\(=>\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)
\(=>35x^2+7x+10x+2=35x^2+49x-5x-7\)
\(=>35x^2+17x+2=35x^2+44x-7\)
\(=>17x+2=44x-7\)
\(=>44x-17x=2+7\)
\(=>27x=9\)
\(=>x=\frac{9}{27}=\frac{1}{3}\)
\(5^x+5^{x+1}+5^{x+2}=775\)
<=>\(5^x.\left(1+5+5^2\right)=775\)
<=>\(5^x.31=775\)
<=>\(5^x=775:31\)
<=>\(5^x=25\)=>\(x=2\)
dễ. k đi mình hứa sẽ làm cho