tìm x:
\(\frac{5^{x}}{5}+6\cdot5^{x}=31\cdot5^{10}\)
Giúp mình với mình cần gấp lắm ạ
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Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
a) \(5^{x+1}-2.5^x=375\)
\(\Rightarrow5^x\left(5-2\right)=375\)
\(\Rightarrow5^x.3=375\)
\(\Rightarrow5^x=125=5^3\)
\(\Rightarrow x=3\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow3^{2\left(x+1\right)}-5.3^{2x}=324\)
\(\Rightarrow3^2\left(3^{x+1}-5.3^x\right)=324\)
\(\Rightarrow9.3^x\left(3-5\right)=324\)
\(\Rightarrow3^x.\left(-2\right)=36\)
\(\Rightarrow3^x=-18=3^2.\left(-2\right)\)(vô lí vì 3x không chia hết cho 2)
c) \(\left(1-x\right)^5=32=2^5\)
\(\Rightarrow1-x=2\)
\(\Rightarrow x=-1\)
d) \(3.5^{2x+1}-3.25^x=300\)
\(\Rightarrow3\left(5^{2x}.5-5^{2x}\right)=300\)
\(\Rightarrow5^{2x}\left(5-1\right)=100\)
\(\Rightarrow5^{2x}.4=100\)
\(\Rightarrow5^{2x}=25=5^2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(a=\frac{6.2.3.4+6.3.4.5+6.4.5.6+...+6.98.99.100}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
thay vào p(x) suy ra a không là nghiệm của đa thức nhé bạn
`a,`
\((- 5) .x + 17 = - 23\)
`\Rightarrow (-5)x = -23 - 17`
`\Rightarrow (-5)x =-40`
`\Rightarrow x = (-40) \div (-5)`
`\Rightarrow x = 8`
Vậy,` x = 8`
`b,`
\(8 + 4x = - 24\)
`\Rightarrow 4x = -24 - 8`
`\Rightarrow 4x = -32`
`\Rightarrow x = -32 \div 4`
`\Rightarrow x = -8`
Vậy, `x = -8`
`c,`
\(32 – 12 + x = -10\)
`\Rightarrow 20 + x = -10`
`\Rightarrow x = -10 - 20`
`\Rightarrow x = -30`
Vậy, `x = -30`
`d,`
\(x – 87 + 13 = - 100\)
`\Rightarrow x - 87 = -100 - 13`
`\Rightarrow x - 87 = -113`
`\Rightarrow x = -113 + 87`
`\Rightarrow x = -26`
Vậy, `x = -26.`
a: \(=\dfrac{2^{13}\cdot5^7\left(2^{17}+5^{20}\right)}{2^{10}\cdot5^7\left(2^{17}+5^{20}\right)}=2^3\)
b: \(M=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)
\(=3^{2\cdot1\cdot13\cdot12}=3^{312}\)
\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)
Ta có:
\(\frac{5^{x}}{5} + 6 \cdot 5^{x} = 31 \cdot 5^{10}\)
Vì
\(\frac{5^{x}}{5} = 5^{x - 1}\)
nên:
\(5^{x - 1} + 6 \cdot 5^{x} = 31 \cdot 5^{10}\)
Đưa \(5^{x - 1}\) ra ngoài:
\(5^{x - 1} \left(\right. 1 + 6 \cdot 5 \left.\right) = 31 \cdot 5^{10}\) \(5^{x - 1} \left(\right. 1 + 30 \left.\right) = 31 \cdot 5^{10}\) \(31 \cdot 5^{x - 1} = 31 \cdot 5^{10}\)
Chia hai vế cho \(31\):
\(5^{x - 1} = 5^{10}\)
Suy ra:
\(x - 1 = 10\)
\(x=11\) \(\)
sao các bn trình bày đc đẹp vậy