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\(M=\frac{-1,2:\left(1\frac{3}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{5}\right):\frac{4}{7}}{\left(5\frac{5}{9}-\frac{9}{4}\right).\frac{36}{17}}+0,6.0,5:\frac{2}{5}\)
\(M=\frac{-1,2:\left(\frac{8}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{0,68:\frac{4}{7}}{\left(\frac{50}{9}-\frac{9}{4}\right).\frac{36}{17}}+0,6.0,5.\frac{5}{2}\)
\(M=\frac{-1.2:2}{0,6}+\frac{1,19}{\frac{119}{36}.\frac{36}{17}}+0.6.0,5.\frac{5}{2}\)
\(M=\frac{-0,6}{0,6}+\frac{1,19}{7}+0,75\)
\(M=-1+0,17+0,75=-0,08=\frac{-2}{25}\)
cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
\(E=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(E=\frac{\frac{4}{5}:\frac{4}{5}:1,25}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right).\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right).\frac{36}{17}}+\frac{6}{5}.\frac{1}{2}.\frac{5}{4}\)
\(E=\frac{1:\frac{5}{4}}{\frac{3}{5}}+\frac{1.\frac{7}{4}}{\frac{119}{36}.\frac{36}{17}}+\frac{3}{4}\)
\(E=\frac{4}{5}.\frac{5}{3}+\frac{\frac{7}{4}}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{7}{4}.\frac{1}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
\(E=\frac{4}{3}+1=\frac{7}{3}\)
Bấm máy tính:
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
E = \(\frac{7}{3}\)
Vậy E = \(\frac{7}{3}\)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)