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Bài giải
a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)
\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\)
\(\Rightarrow\text{ }x+5=0\)
\(x=0-5\)
\(=-5\)
\(\left(-7\right)^x=\frac{1}{49}\)
=>\(\left(-7\right)^x=\frac{1}{\left(-7\right)^2}\)
=>\(\left(-7\right)^x=\left(-7\right)^{-2}\)
=>x=-2
câu thứ 2 sai đề hay sao á
a ) \(\left(-7\right)^x=\frac{1}{49}\)
\(\left(-7\right)^x=\frac{1}{\left(-7\right)^2}\)
\(\left(-7\right)^x=\frac{1}{\left(-7\right)^{-2}}\)
=> \(x=-2\)
b ) \(\left(-2\right)^x=-0,125\)
\(\left(-2\right)^x=\left(-2\right)^{-3}\)
\(\Rightarrow x=-3\)
<=>(2/7x+1)^2=(4/7)^2
=> 2/7x+1=4/7 hoặc 2/7x+1=-4/7
<=>x=-3/2 hoặc x=-11/2
\(\frac{x}{2}-\left(\frac{3}{5}x-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
\(\Rightarrow\frac{5x-6x+26+14+7x}{10}=0\Rightarrow6x+40=0\Rightarrow x=-\frac{20}{3}\)
a. \(25.5^3.\frac{1}{625}.5^2=5^2.5^3.\frac{1}{5^4}.5^2=\frac{5^7}{5^4}=5^3\)
b. \(4.32:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{2^4}=\frac{2^4}{2^4}=1\)
c. \(5^2.3^5.\left(\frac{3}{5}\right)^2=5^2.3^5.3^2.\frac{1}{5^2}==\frac{5^2}{5^2}.3^7=3^7\)
d. \(\left(\frac{1}{7}\right)^2.\frac{1}{7}.49^2=\frac{1}{7^3}.7^4=\frac{7^4}{7^3}=7\)
\(\frac{\left(-7\right)^{^{x-1}}}{49}\)=\(\frac{-49}{49}\)
=>\(\left(-7\right)^{^{x-1}}\)=-49
\(\left(-7\right)^{^{x-1}}\)=\(\left(-7\right)^2\)
=> x-1 = 2
x = 2+1
x = 3