Bài 1: Thực hiện phép tính a) \(\frac{3}{5}+\left(-\frac{2}{3}\right)-\left(-\frac{5}{2}\right)\) b) \(2\frac{2}{9}:1\frac{1}{9}-\frac{46}{5}:4\frac{3}{5}\) c) \(\left(\frac{2}{3}\right)^{^2}+\frac{5}{9}+\left(\frac{2}{3}:\frac{-2}{3}\right)\) d) \(\frac{5^2\cdot5^3}{\left(-5\right)^4}\)
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\(A=9-\frac{3}{5}+\frac{2}{3}-7-\frac{7}{5}+\frac{3}{2}-3+\frac{9}{5}-\frac{5}{2}\)
\(=\left(9-7-3\right)+\left(\frac{9}{5}-\frac{7}{5}-\frac{3}{5}\right)+\left(\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2-\frac{1}{5}=-\frac{11}{5}\)
a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)
\(=10:\left(\frac{-2}{3}\right)\)
\(=-15\)
b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\frac{-8}{27}+\frac{1}{10}\)
\(=\frac{-8}{3}+\frac{1}{10}\)
\(=\frac{-77}{30}\)
c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)
\(=\frac{-1}{3}\)
\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)
\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)
\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)
\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)
\(=-15\)
\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)
\(=-\frac{8}{3}+\frac{1}{10}\)
\(=-\frac{77}{30}\)
\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)
\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)
\(=\frac{1}{3}\)
Theo đề ta có:
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)\)\(-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\)\(\frac{1}{5}-\frac{2}{9}\)-\(2+\frac{1}{23}+2+\frac{3}{5}+\frac{5}{6}-8+\frac{2}{3}-\frac{1}{18}\)
=\(\left(5+2-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{5}{6}-\frac{2}{3}+\frac{1}{18}\right)+\frac{1}{23}\)
= -1 +\(\frac{4}{5}\)\(-\frac{-11}{9}\)+\(\frac{1}{23}\)
= -1 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)
= \(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)
= \(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)
= \(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)
= \(-\frac{38}{15}+\frac{1}{23}\)
= \(-\frac{859}{345}\)
a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)
\(=8+3.1+4:\frac{1}{2}\)
\(=8+3+8=19\)
b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)
d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)
\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left(\frac{-3}{22}\right)+\frac{5}{9}:\left(\frac{-3}{5}\right)\)
\(=\frac{5}{9}.\left(-\frac{22}{3}\right)+\frac{5}{9}.\left(-\frac{5}{3}\right)\)
\(=\frac{5}{9}.\left[-\frac{22}{3}+\left(-\frac{5}{3}\right)\right]\)
\(=\frac{5}{9}.\left(-9\right)\)
\(=-5\)
\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left[\left(\frac{1}{11}-\frac{5}{22}\right)+\left(\frac{1}{15}-\frac{2}{3}\right)\right]\)
\(=\frac{5}{9}:\left[-\frac{3}{22}-\frac{3}{5}\right]\)
\(=\frac{5}{9}:\frac{-81}{110}=\frac{5}{9}.\frac{-110}{81}\)
\(=-\frac{550}{729}\)
a. \(\frac{3}{5}+\left(-\frac{2}{3}\right)-\left(-\frac{5}{2}\right)=\frac{3}{5}-\frac{2}{3}+\frac{5}{2}=\frac{73}{30}\)
b. \(2\frac{2}{9}:1\frac{1}{9}-\frac{46}{5}:4\frac{3}{5}=\frac{20}{9}:\frac{10}{9}-\frac{46}{5}:\frac{23}{5}=\frac{20}{9}.\frac{9}{10}-\frac{46}{5}.\frac{5}{23}=2-2=0\)
c. \(\left(\frac{2}{3}\right)^2+\frac{5}{9}+\left(\frac{2}{3}:\frac{-2}{3}\right)=\frac{4}{9}+\frac{5}{9}+\left(\frac{2}{3}.\frac{-3}{2}\right)=1+\left(-1\right)=0\)
d. \(\frac{5^2.5^3}{\left(-5\right)^4}=\frac{5^5}{5^4}=5\)
a, \(\frac{3}{5}+\left(-\frac{2}{3}\right)-\left(-\frac{5}{2}\right)=\frac{3}{5}-\frac{2}{3}+\frac{5}{2}=\frac{73}{30}\)
b, \(2\frac{2}{9}:1\frac{1}{9}-\frac{46}{5}:4\frac{3}{5}=\frac{20}{9}:\frac{10}{9}-\frac{46}{5}:\frac{23}{5}\)
\(=2-2=0\)
c, \(\left(\frac{2}{3}\right)^2+\frac{5}{9}+\left(\frac{2}{3}:\frac{-2}{3}\right)=\frac{4}{9}+\frac{5}{9}-1=0\)
d, \(\frac{5^2.5^3}{\left(-5\right)^4}=\frac{5.5.5.5.5}{5.5.5.5}=5\)Vì \(\left(-5\right)^4=5^4\)