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b)\(\left(2\frac{1}{3}+3\frac{1}{2}\right).0,2+25\%=\left(\frac{7}{3}+\frac{7}{2}\right).\frac{1}{5}+\frac{1}{4}=\left(\frac{14}{6}+\frac{21}{6}\right).\frac{1}{5}+\frac{1}{4}=\frac{35}{6}.\frac{1}{5}+\frac{1}{4}=\frac{7}{6}+\frac{1}{4}=\frac{14}{12}+\frac{3}{12}=\frac{17}{12}=1\frac{5}{12}\)
1a) (x-2)(x+1)= 0
Suy ra \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\)Suy ra \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
a; - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - (- \(\dfrac{1}{6}\)) + (- \(\dfrac{2}{5}\))
= - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) + \(\dfrac{1}{6}\) - \(\dfrac{2}{5}\)
= \(-\dfrac{40}{60}\) + \(\dfrac{45}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{5}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{15}{60}\) - \(\dfrac{24}{60}\)
= - \(\dfrac{3}{20}\)
b; (- \(\dfrac{2}{3}\)) + (- \(\dfrac{1}{5}\)) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) - \(\dfrac{-7}{10}\)
= - \(\dfrac{2}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{10}\)
= - \(\dfrac{40}{60}\) - \(\dfrac{12}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{52}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{7}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{57}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{1}{4}\)
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https://www.youtube.com/channel/UCT23clmdY5azigRNMRDxGfw
a) \(\left(x^2+5\right).\left(x^2-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+5=0\\x^2-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-5\left(vl\right)\\x^2=25\end{cases}\Rightarrow}\orbr{\begin{cases}\\x=\pm5\end{cases}}}\)
b) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)
\(\Rightarrow\left(x^2-5\right)\)và \(\left(x^2-25\right)\)trái dấu
Vì \(\left(x^2-5\right)>\left(x^2-25\right)\)
\(\Rightarrow\hept{\begin{cases}x^2-5>0\\x^2-25< 25\end{cases}\Rightarrow\hept{\begin{cases}x^2>5\\x^2< 50\end{cases}}}\)
\(\Rightarrow5< x^2< 50\)
\(\Rightarrow x^2\in\left\{0;1;4;9;16;25;36;49\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5;\pm6;\pm7\right\}\)
c) \(\left(x-2\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
các câu còn lại lm tương tự nhé!! hok tốt!!
\(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
\(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{5^4\cdot4^4\cdot5^4}{5^5\cdot5^5.4^5}=\frac{1}{100}\)
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2=3-1+\frac{1}{8}=\frac{17}{8}\)