\(\frac{12}{6}\)x\(\frac{4}{5}\)' 4\(^5\)x4
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\(\frac{2x}{3}-\frac{x}{4}=\frac{5}{6}\)
\(\Leftrightarrow\frac{2x}{3}.12-\frac{x}{4}.12=\frac{5}{6}.12\)
<=> 8x - 3x = 10
<=> 5x = 10
<=> x = 10 : 5
<=> x = 2
=> x = 2
1) \(\frac{x-3}{2}+\frac{4x+1}{3}=\frac{2x-7}{6}\)
<=> 3(x - 3) + 2(4x + 1) = 2x - 7
<=> 3x - 9 + 8x + 2 = 2x - 7
<=> 11x - 7 = 2x - 7
<=> 11x - 7 - 2x = -7
<=> 9x - 7 = -7
<=> 9x = -7 + 7
<=> 9x = 0
<=> x = 0
Ta có \(81.\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{3}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right)\)
\(=81.\left(\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right)\)
\(=81.\left(\frac{12}{4}:\frac{5}{6}\right)\)
\(=81.\frac{18}{5}\)
\(=291,6\)
\(81\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{159}+\frac{6}{91}}\right)\)
\(=81\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{2}{169}+\frac{1}{91}\right)}\right)\)
\(=81\left(3\div\frac{5}{6}\right)\)
\(=81.\frac{18}{5}\)
\(=\frac{1458}{5}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{241864704+1209323520}{2176782336-362797056}=\frac{4}{5}\)
\(\frac{12}{6}.\frac{4}{5}.4^5.4\)
\(=\left(\frac{12}{6}.\frac{4}{5}\right).4^5.4\)
\(=\frac{8}{5}.4^5.4\)
\(=\frac{8}{5}.1024.4\)
\(=1638,4.4\)
\(=6553.6\)