Cho A = \(\frac{5}{4.3}\) + \(\frac{14}{3.23}\) + \(\frac{16}{23.13}\) + \(\frac{7}{13.46}\) . Tìm A
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\(A=\frac{81^4.3^{10}.27^5:3^{12}}{3^{18}:9^3.243^2}=\frac{3^{16}.3^{10}.3^{15}:3^{12}}{3^{18}:3^6.3^{10}}=\frac{3^{29}}{3^{22}}=3^7\)
\(B=\frac{2.55^2-9.55^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{2.55^2-9.55^2.55^{19}}{25^{10}}:\frac{5\left(21.7^{14}-19.7^{14}\right)}{7.7^{15}+3.7^{15}}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}:\frac{5.7^{14}.2}{7^{15}.10}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}.\frac{7^{15}.10}{5.7^{14}.2}\)Chịu ==
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.