(2/3x3/5)(3/-2-10/3)=2/5
Tìm x biết :
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=6/15x(-9/6-20/6)
=6/15x[-(9/6+20/6)]
=6/15x(-29/6)
=-174/90
=-29/15
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
\(\left(\frac{-2}{3x}-\frac{3}{5}\right)\left(\frac{3}{-2}-\frac{10}{3}\right)\)
\(=\left[-\left(\frac{2}{3x}+\frac{3}{5}\right)\right]\left[-\left(\frac{3}{2}+\frac{10}{3}\right)\right]\)
\(=\left(\frac{2}{3x}+\frac{3}{5}\right)\left(\frac{3}{2}+\frac{10}{3}\right)\)
\(=\left(\frac{10}{15x}+\frac{9x}{15x}\right)\left(\frac{9}{6}+\frac{20}{6}\right)\)
\(=\frac{10+9x}{15x}.\frac{9+20}{6}\)
\(=\frac{29.\left(10+9x\right)}{90}\)
\(\frac{5}{3}x-\frac{2}{5}x=\frac{19}{10}\)
\(\left(\frac{5}{3}-\frac{2}{5}\right)x=\frac{19}{10}\)
\(\frac{19}{15}x=\frac{19}{10}\)
\(x=\frac{19}{30}\)
\(\frac{5}{3}x-\frac{2}{5}x=\frac{19}{10}\)
(5/3 - 2/5)x = 19/10
19/15x = 19/10
x = 19/30
có phải nhu thế này không
\((\frac{2}{3}x\frac{3}{5})(\frac{3}{-2}-\frac{10}{3})=\frac{2}{5}\)
\(\left(\frac{2}{5}x\right)\left(\frac{-29}{6}\right)=\frac{2}{5}\)
\(\frac{2}{5}x=\frac{2}{5}.\frac{-6}{29}\)
\(x=\frac{\frac{2}{5}.\frac{-6}{29}}{\frac{2}{5}}=\frac{-6}{29}\)