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Bài 4:
Điện trở tương đương: \(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{40.60}{40+60}=24\left(\Omega\right)\)
Do mắc song song nên \(U=U_1=U_2=I.R_{tđ}=0,5.24=12\left(V\right)\)
Cường độ dòng điện qua mỗi điện trở:
\(\left\{{}\begin{matrix}I_1=\dfrac{U_1}{R_1}=\dfrac{12}{40}=0,3\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{12}{60}=0,2\left(A\right)\end{matrix}\right.\)
\(P=U.I=12.0,5=6\left(W\right)\)
\(A=P.t=6.1.60.60=21600\left(J\right)\)
Bài 5:
Điện trở tương đương của cả mạch: \(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{3.6}{3+6}=2\left(\Omega\right)\)
Do mắc song song nên \(U=U_1=U_2=18V\)
Cường độ dòng điện qua mỗi điện trở và cả mạch:
\(\left\{{}\begin{matrix}I=\dfrac{U}{R_{tđ}}=\dfrac{18}{2}=9\left(A\right)\\I_1=\dfrac{U_1}{R_1}=\dfrac{18}{3}=6\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{18}{6}=3\left(A\right)\end{matrix}\right.\)
Tiết diện của dây: \(R_2=\rho_2.\dfrac{l_2}{S_2}\Rightarrow S_2=\dfrac{\rho_2.l_2}{R_2}=\dfrac{1,6.10^{-6}.10}{6}\approx2,67.10^{-6}\left(m^2\right)\)
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a) Với x>0,x\(\ne\)9
\(Q=\left(\dfrac{1}{x-3\sqrt{x}}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-3\right)^2}=\left(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\right).\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
b)Với x>0,x\(\ne\)9
\(Q< 0< =>\dfrac{\sqrt{x}-3}{\sqrt{x}}< 0\)
\(< =>\sqrt{x}-3< 0\left(Vì\sqrt{x}>0\right)\)
\(< =>\sqrt{x}< 3\)
\(< =>x< 9\)
Kết hợp với ĐKXĐ ta được
0<x<9
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\(\sqrt{\left(24+8\sqrt{5}\right)^2}-\sqrt{\left(9-4\sqrt{5}\right)^2}=24+8\sqrt{5}-9+4\sqrt{5}=15+12\sqrt{5}\)
\(\sqrt{\left(17-12\sqrt{2}\right)^2}+\sqrt{\left(9+4\sqrt{2}\right)^2}=17-12\sqrt{2}+9+4\sqrt{2}=26-8\sqrt{2}\)
\(\sqrt{\left(6-4\sqrt{2}\right)^2}+\sqrt{\left(22-12\sqrt{2}\right)^2}=6-4\sqrt{2}+22-12\sqrt{2}=28-16\sqrt{2}\)
\(ô,\\ \Rightarrow24+8\sqrt{5}-\sqrt{\left(9-4\sqrt{5}\right)^2}\\ \Rightarrow24+8\sqrt{5}-\left(9-4-\sqrt{5}\right)\\ \Rightarrow24+8\sqrt{5}-9+4\sqrt{5}\\ \Rightarrow15+8\sqrt{5}+4\sqrt{5}\\ \Rightarrow15+12\sqrt{5}\)
\(ơ,\\ g\left(17-12\sqrt{2}\right)+\sqrt{\left(9+4\sqrt{2}\right)^2}\\ \Rightarrow g\left(17-12\sqrt{2}\right)+\sqrt{\left(9+4+\sqrt{2}\right)^2}\\ \Rightarrow\left(17-12\sqrt{2}\right)g+\sqrt{\left(9+4\sqrt{2}\right)^2}\\ \Rightarrow\left(17-12\sqrt{2}\right)g+9+4\sqrt{2}\)
\(u,\\ 6-4\sqrt{2}+\sqrt{\left(22-12\sqrt{2}\right)}^2\\ \Rightarrow6-4\sqrt{2}+22-12\sqrt{2}\\ \Rightarrow28-4\sqrt{2}-12\sqrt{2}\\ \Rightarrow28-16\sqrt{2}\)
a)
\(0,3-\dfrac{8}{3}:\dfrac{4}{3}\cdot\dfrac{1}{5}+1\\ =\dfrac{3}{10}-\dfrac{8}{3}\cdot\dfrac{3}{4}\cdot\dfrac{1}{5}+1\\ =\dfrac{3}{10}-2\cdot\dfrac{1}{5}+1\\ =\dfrac{3}{10}-\dfrac{2}{5}+1\\ =\dfrac{9}{10}\)
b)
\(\left(-\dfrac{1}{2}\right)^2-\dfrac{5}{8}:\left(0,5\right)^3-\dfrac{5}{3}\cdot\left(-6\right)\\ =\dfrac{1}{4}-\dfrac{5}{8}:\dfrac{1}{8}+\dfrac{5}{3}\cdot6\\ =\dfrac{1}{4}-\dfrac{5}{8}\cdot8+10\\ =\dfrac{1}{4}-5+10\\ =\dfrac{21}{4}\)
c)
\(2+4:\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\\ =2+4:\dfrac{1}{2}\cdot\left(-2,25\right)\\ =2+8\cdot\left(-2,25\right)\\ =2-18\\ =-16\)
d)
\(\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\\ =\left(2+\dfrac{5}{6}+1+\dfrac{4}{9}\right):\left(10+\dfrac{1}{12}-9-\dfrac{1}{2}\right)\\ =\left(3+\dfrac{23}{18}\right):\left(1-\dfrac{5}{12}\right)\\ =\dfrac{77}{18}:\dfrac{7}{12}\\ =\dfrac{22}{3}\)
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