{36x4-4x[(2^4x5+2)-7x11]^2}:4-2018^0
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\(\left[36x4x\left(82-7x11\right)^2\right]:4-2016^0\)
\(=\left[144x\left(82-77\right)^2\right]:4-1\)
\(=\left(144x5^2\right):4-1\)
\(=\left(144x25\right):4-1\)
\(=3600:4-1\)
\(=900-1\)
\(=899\)
[ 36.4.( 82 - 7.11 )2 ] : 4 - 20160
= [ 36.4.52 ] : 4 - 20160
= [ 36.4.25 ] : 4 - 20160
= 3600 : 4 -1
= 900 - 1
= 899
a) \(\left(4x^2-2\right)^2=\frac{196}{81}\)
<=> \(2^2\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(4\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(\left(2x^2-1\right)^2=\frac{196}{81}:4\)
<=> \(\left(2x^2-1\right)^2=\frac{49}{81}\)
<=> \(2x^2-1=\pm\sqrt{\frac{49}{81}}\)
<=> \(2x^2-1=\pm\frac{7}{9}\)
<=> \(\orbr{\begin{cases}2x^2-1=\frac{7}{9}\\2x^2-1=-\frac{7}{9}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
Ta có: \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^6}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^6}}{2}\)