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Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ góc trái khung soạn thảo)
1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)
2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)
\(2^x=2.2^8=2^9;x=9\)
4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)
\(\Leftrightarrow3^5.x=3^5.5;x=5\)
Dấu "^" của cậu nghĩa là phân số?
\(a,-1,75-\left(-\dfrac{1}{9}-2\dfrac{1}{18}\right)=-\dfrac{7}{4}-\left(-\dfrac{2}{18}-\dfrac{37}{18}\right)=-\dfrac{7}{4}+\dfrac{13}{6}=\dfrac{5}{12}\)
\(b,-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\dfrac{55}{24}=-\dfrac{19}{8}\)
\(c,-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{30}\right)=-\dfrac{5}{6}+\dfrac{41}{120}=-\dfrac{59}{120}\)
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left[\left(\frac{2}{3}-\frac{4}{7}\right)+\left(-\frac{8}{7}+\frac{1}{3}\right)\right]:\frac{5}{9}\)
\(=\left(\frac{2}{3}-\frac{4}{7}-\frac{8}{7}+\frac{1}{3}\right)\cdot\frac{9}{5}\)
\(=\left(1-\frac{12}{7}\right)\cdot\frac{9}{5}\)
\(=-\frac{5}{7}\cdot\frac{9}{5}\)
\(-\frac{9}{7}\)
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left(\frac{14}{21}-\frac{12}{21}\right):\frac{5}{9}+\left(-\frac{24}{21}+\frac{7}{21}\right):\frac{5}{9}\)
\(=\frac{2}{21}:\frac{5}{9}+\frac{-17}{21}:\frac{5}{9}\)
\(=\left(\frac{2}{21}+\frac{-17}{21}\right):\frac{5}{9}\)
\(=\frac{-15}{21}:\frac{5}{9}\)
\(=\frac{-15}{21}.\frac{9}{5}\)
\(=\frac{-9}{7}\)
\(=\frac{99}{35}\)
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
a) \(\frac{-5}{8}\cdot\frac{11}{3}+\frac{-5}{8}\cdot\frac{1}{3}=-\frac{5}{8}\left(\frac{11}{3}+\frac{1}{3}\right)=-\frac{5}{8}\cdot4=-\frac{5}{2}\cdot1=-\frac{5}{2}\)
b) \(\frac{2}{3}+\frac{3}{4}\cdot\frac{9}{5}=\frac{2}{3}+\frac{27}{20}=\frac{121}{60}\)
c) Tương tự câu a
d) \(\frac{1}{7}\cdot\frac{3}{8}+\frac{1}{7}\cdot\frac{5}{8}=\frac{1}{7}\left(\frac{3}{8}+\frac{5}{8}\right)=\frac{1}{7}\cdot1=\frac{1}{7}\)
\(a,\frac{-5}{8}.\frac{11}{3}+\frac{-5}{8}.\frac{1}{3}\)
\(=\frac{-5}{8}\left(\frac{11}{3}+\frac{1}{3}\right)\)
\(=\frac{-5}{8}.4\)
\(=\frac{-5}{2}\)
\(b,\frac{2}{3}+\frac{3}{4}.\frac{9}{5}\)
\(=\frac{2}{3}+\frac{27}{20}\)
\(=\frac{40}{60}+\frac{81}{60}\)
\(=\frac{121}{60}\)
\(c,\frac{-5}{7}.\frac{4}{9}-\frac{5}{9}.\frac{5}{7}\)
\(=\frac{-5}{7}\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=\frac{-5}{7}.1\)
\(=\frac{-5}{7}\)
\(d,\frac{1}{7}.\frac{3}{8}+\frac{1}{7}.\frac{5}{8}\)
\(=\frac{1}{7}\left(\frac{3}{8}+\frac{5}{8}\right)\)
\(=\frac{1}{7}.1\)
\(=\frac{1}{7}\)
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