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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
=>\(x=-\dfrac{5}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
\(5^2.7^2.11.\left(77x+5\right)=0\)
\(77x+5=0\)
\(77x=-5\)
\(x=\dfrac{-5}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
\(\Leftrightarrow5^2.7^2.11\left(7.11.x+5\right)=0\)
\(\Leftrightarrow77x+5=0\)
\(\Leftrightarrow77x=-5\)
\(\Leftrightarrow x=-\frac{5}{77}\)
( 3x - 24 ) . 75 = 2.76 .1/20170
( 3x - 24 ) . 75 =235298
( 3x - 24 ) = 235298 : 75
( 3x - 24 ) =14
3x = 14 + 24
3x = 30
x = 0
dung 100%
\(\left(3x-2^4\right).7^5=2.7^6.\frac{1}{2017^0}\)
\(\Leftrightarrow\left(3x-16\right).7^5=2.7^6.1\)
\(\Leftrightarrow3x-16=\frac{2.7^6}{7^5}\)
\(\Leftrightarrow3x-16=2.7\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\)
a) (3x-24) = 2.74:73
=> 3x-24 = 2.7
=> 3x-16 = 14
=> 3x = 14+16
=> 3x = 30
=> x = 30:3
Vậy x = 10
b) x - [42 + (-28)] = -8
=> x - 14 = -8
=> x = -8 + 14
Vậy x = 6
c) l x-3 l = l 5 l + l -7 l
=> l x-3 l = 5+7
=> l x-3 l = 12
=> x-3 = 12 hay x-3 = -12
=> x = 12+3 hay x = -12+3
Vậy x = 15 hay x = -9
d) mình k biết
\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)
<=>\(\left(\frac{2}{3}\right)^{\frac{n}{2}}=\left(\frac{2}{3}\right)^5\)
<=>\(\frac{n}{2}=5\)
<=>n=10
\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)
\(\Rightarrow\left(\frac{2}{3}\right)^{2n}=\left(\frac{2}{3}\right)^5\)
\(\Rightarrow2n=5\Rightarrow n=\frac{5}{2}\)
Vậy n = 5/2
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