\(\frac{2}{9}\)+\(\frac{-2}{9}\)x\(\frac{2}{9}\): \(\frac{14}{81}\)
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=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)
=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)
=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)
=>x-3=0
=>x=3
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.
\(\left(\frac{81}{12}-\frac{9}{4}\right).\left(\frac{2}{9}\right)^2-\frac{2}{9}\)
\(=\left(\frac{27}{4}-\frac{9}{4}\right).\frac{4}{81}-\frac{2}{9}\)
\(=\left(\frac{27}{4}-\frac{9}{4}\right).\frac{4}{81}-\frac{2}{9}\)
\(=\frac{9}{2}.\frac{4}{81}-\frac{2}{9}\)
\(=\frac{9.4}{2.81}-\frac{2}{9}\)
\(=\frac{1.2}{1.9}-\frac{2}{9}\)
\(=\frac{2}{9}-\frac{2}{9}=0\)
\(\left(\frac{81}{12}-\frac{9}{4}\right).\left(\frac{2}{9}\right)^2-\frac{2}{9}\)
\(=\left(\frac{27}{4}-\frac{9}{4}\right).\left(\frac{2}{9}.\frac{2}{9}\right)-\frac{2}{9}\)
\(=\frac{9}{2}.\frac{4}{81}-\frac{2}{9}\)
\(=\frac{9.4}{2.81}-\frac{2}{9}=\frac{2}{9}-\frac{2}{9}=0\)
Vậy \(\left(\frac{81}{12}-\frac{9}{4}\right).\left(\frac{2}{9}\right)^2-\frac{2}{9}=0\)
1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)
\(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)
( . là nhân nha)
\(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)
\(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)
\(x=\frac{113}{8}\)
( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{49}{81}=\frac{56}{81}\)
\(4y=\frac{7}{81}\)
y = 7/81:4
y = 7/324
\(A=\frac{9}{8}-\frac{8}{9}+\frac{3}{25}+\frac{1}{4}-\frac{5}{16}+\frac{19}{25}-\frac{1}{9}+\frac{2}{25}-\frac{1}{81}\)
\(A=\left(\frac{9}{8}+\frac{1}{4}-\frac{5}{16}\right)-\left(\frac{8}{9}+\frac{1}{9}-\frac{1}{81}\right)+\left(\frac{3}{25}+\frac{19}{25}+\frac{2}{25}\right)\)
\(A=\frac{17}{16}-\frac{80}{81}+\frac{24}{25}\)
\(A=\frac{33529}{32400}\)
= -4/3 nha
\(\frac{2}{9}\)+ \(\frac{-2}{9}\) x\(\frac{2}{9}\):\(\frac{14}{81}\)
= \(\left(\frac{2}{9}+\frac{-2}{9}\right)\) x \(\frac{2}{9}\): \(\frac{14}{81}\)
= 0 x \(\frac{2}{9}\): \(\frac{14}{81}\)
= 0
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