Tính giá trị của biểu thức sau: 3280-(3^2*7^3-2^3*49)
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\(\frac{\left(\frac{3}{15}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{5\frac{1}{3}+\frac{2}{5}}\)
\(=\frac{\left(\frac{12}{60}+\frac{15}{60}+\frac{21}{60}\right)\times\frac{17}{49}}{\frac{16}{3}\times\frac{2}{5}}\)
\(=\frac{\frac{48}{60}\times\frac{17}{49}}{\frac{80}{15}+\frac{6}{15}}\)
\(=\frac{\frac{816}{2940}}{\frac{86}{15}}\)
\(=\frac{816}{2940}:\frac{86}{15}\)
\(=\frac{816}{2940}\times\frac{15}{86}\)
\(=\frac{68}{245}\times\frac{15}{86}\)
\(=\frac{102}{2107}\)
a: \(B=\left(4\cdot2^5\right):\left(2^3\cdot\frac{1}{16}\right)\)
\(=\left(4\cdot32\right):\left(\frac{8}{16}\right)\)
\(=128:\frac12=128\cdot2=256\)
b: \(B=3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^3}\)
\(=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=\frac{3^8}{3^8}\cdot3^2=3^2=9\)
c: \(D=\left\lbrace\left(0,1\right)^2\right\rbrace^0+\left\lbrack\left(\frac17\right)^1\right\rbrack^2:\frac{1}{49}\cdot\left\lbrack\left(2^2\right)^3:2^5\right\rbrack\)
\(=1+\left(\frac17\right)^2\cdot49\cdot2^6:2^5\)
\(=1+49\cdot\frac{1}{49}\cdot2=1+2=3\)
d: \(C=\left(-0,5\right)^5:\left(-0,5\right)^3-\left(\frac{17}{2}\right)^7:\left(\frac{17}{2}\right)^6\)
\(=\left(-0,5\right)^{5-3}-\left(\frac{17}{2}\right)\)
\(=\left(-0,5\right)^2-\frac{17}{2}=0,25-\frac{17}{2}=\frac14-\frac{34}{4}=-\frac{33}{4}\)
(140 x \(\frac{7}{3}\) - 138 x \(\frac{5}{12}\)) : \(\frac{49}{6}\)=
=\(<\frac{980}{3}-\frac{115}{2}>:\frac{49}{6}\)
= (\(\frac{1960}{6}-\frac{345}{6}\) ) :\(\frac{49}{6}\)
=\(\frac{1615}{6}:\frac{49}{6}\)
=\(\frac{1615}{49}\)
Ta có :\(A=3280-\left(3^2.7^2-2^3.49\right)=3280-\left(3^2.7.7^2-2^3.7^2\right)\)
\(=3280-7^2\left(72-8\right)\)
\(=3280-49.82=3280-3136=144\)
\(3280-\left(3^2.7^3-2^3.49\right)\)
\(=3280-\left(9.7^3-8.7^2\right)\)
\(=3280-7^2\left(9.7-8\right)=3280-49.55=3280-2695=585\)