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ta có \(5^2.7^3.11^2.x+5^3.7^2.11=0< =>5^2.7^2.11\left(77x+1\right)=0\)
<=> \(77x+1=0< =>x=-\frac{1}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
=>\(5^2.7^2.11.\left(77x+5\right)=0\)
=>\(77x+5=0\)
=>77x=-5
=>\(x=-\dfrac{5}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
\(5^2.7^2.11.\left(77x+5\right)=0\)
\(77x+5=0\)
\(77x=-5\)
\(x=\dfrac{-5}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
\(\Leftrightarrow5^2.7^2.11\left(7.11.x+5\right)=0\)
\(\Leftrightarrow77x+5=0\)
\(\Leftrightarrow77x=-5\)
\(\Leftrightarrow x=-\frac{5}{77}\)
`5x - 20 - x - 2 = 2x - 6`
`<=> 4x - 22 = 2x - 6`
`<=> 2x = 16`
`<=>x=8`
Ta có: 52.73.112.x - 52.72.114 = 0
=> 52.72.112(7x - 112) = 0
=> 7x - 121 = 0
=> 7x = 121
=> x = 121 : 7
=> x = 121/7
Câu 2:
\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)
\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)
\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)
<=>5^2.7^3x+5^3.7^2.11=125(7x+55)
=>1225(7x+55)=0
=>8575x+67375=0
=>8575x=-67375
=>1225.7x=1225.(-55)
=>7x=-55
=>x=\(\frac{-55}{7}\)
x=\(\frac{-55}{7}\)