câu 2 câu 3 câu 4 câu 5
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Câu 1:
\(\dfrac{-5}{6}:\dfrac{3}{13}=\dfrac{-5}{6}.\dfrac{13}{3}=\dfrac{-5.13}{6.3}=\dfrac{-65}{18}\)
Câu 2:
\(\dfrac{5}{9}:\dfrac{5}{-3}=\dfrac{5}{9}.\dfrac{-3}{5}=\dfrac{5.-3}{9.5}=\dfrac{-15}{45}=\dfrac{-1}{3}\)
Câu 3:
\(\left(-15\right):\dfrac{3}{2}=\left(-15\right).\dfrac{2}{3}=\dfrac{-15.2}{3}=\dfrac{-30}{3}=-10\)
Câu 4:
\(\dfrac{3}{4}:\left(-9\right)=\dfrac{3}{4}.\dfrac{-1}{9}=\dfrac{3.-1}{4.9}=\dfrac{-3}{36}=\dfrac{-1}{12}\)
Câu 1
\(-\dfrac{5}{6}:\dfrac{3}{13}=-\dfrac{5}{6}.\dfrac{13}{3}=\dfrac{-5.13}{6.3}=-\dfrac{65}{18}\)
Câu 2
\(2:\dfrac{5}{9}:\dfrac{5}{-3}=2.\dfrac{9}{5}.\dfrac{-3}{5}=\dfrac{2.9.\left(-3\right)}{5.5}=-\dfrac{54}{25}\)
Câu 3
\(4:\dfrac{3}{2}:\left(-9\right)=4.\dfrac{2}{3}.\dfrac{1}{-9}=\dfrac{4.2.1}{-9}=-\dfrac{8}{9}\)
Câu 2.
\(R_{13}=\dfrac{R_1\cdot R_3}{R_1+R_3}=\dfrac{30\cdot60}{30+60}=20\Omega\)
\(R_{tđ}=R_2+R_{13}=45+20=65\Omega\)
\(I_m=\dfrac{U}{R}=\dfrac{130}{65}=2A\)
\(I_{13}=I_m=2A\)
\(U_{13}=U-U_2=U-I_2\cdot R_2=130-2\cdot45=40V\)
\(R_1//R_3\Rightarrow U_1=U_3=U_{13}=40V\)
\(I_3=\dfrac{U_3}{R_3}=\dfrac{40}{60}=\dfrac{2}{3}A\)
Câu 3.
\(R_1//R_2\)
\(\Rightarrow R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{10\cdot30}{10+30}=7,5\Omega\)
\(U_1=U_2=U_m=15V\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{15}{10}=1,5A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{15}{30}=0,5A\)
\(I_m=I_1+I_2=1,5+0,5=2A\)
Câu 1:
\(\dfrac{2}{5}-\dfrac{1}{4}+\dfrac{3}{10}=\dfrac{8}{20}-\dfrac{5}{20}+\dfrac{6}{20}=\dfrac{8-5+6}{20}=\dfrac{9}{20}\)
Câu 2:
\(\dfrac{-2}{5}:\left(1-\dfrac{1}{10}\right)=\dfrac{-2}{5}:\dfrac{9}{10}=\dfrac{-2}{5}.\dfrac{10}{9}=\dfrac{-2.10}{5.9}=\dfrac{-20}{45}=\dfrac{-4}{9}\)
Câu 3:
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}=\dfrac{7}{18}+\dfrac{1}{5}=\dfrac{53}{90}\)
Câu 4:
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)
\(=\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=\dfrac{2}{7}.1\)
\(=\dfrac{2}{7}\)
Câu 1
\(\dfrac{2}{5}\)-\(\dfrac{1}{4}\)+\(\dfrac{3}{10}\)= \(\dfrac{8}{20}\)-\(\dfrac{5}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{3}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{9}{20}\)
Câu 2
-\(\dfrac{2}{5}\):(1-\(\dfrac{1}{10}\))= -\(\dfrac{2}{5}\):\(\dfrac{9}{10}\)=-\(\dfrac{2}{5}\).\(\dfrac{10}{9}\)=-\(\dfrac{4}{9}\)
Câu 3
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}\)= \(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}.\dfrac{14}{5}\)=\(\dfrac{7.4}{4.2.9}+\dfrac{1.14}{14.5}\)=\(\dfrac{7}{18}+\dfrac{1}{5}\)=\(\dfrac{35}{90}+\dfrac{18}{90}\)=\(\dfrac{53}{90}\)
Câu 4
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)=\(\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)=\(\dfrac{2}{7}.1\)=\(\dfrac{2}{7}\)
Bài 3:
a: \(A=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}:\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-4\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-4\right)\left(x+3\right)}\cdot\dfrac{x+2}{x-1}=\dfrac{x+2}{x-1}\)
b: Để A=3/2 thì 3(x-1)=2(x+2)
=>3x-3=2x+4
=>x=7(nhận)
2 they have written to us the second time.
3 we have learn french for 2 years by mr smith.
4 properly theydidn't look after the children.
xong nha bn
<Hok tốt>