25.

\(R=\dfrac{\left(R1+R2\right)R3}{R1+R2+R3}=\dfrac{\left(2+4\right)6}{2+4+6}=3\Omega\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{6}{3}=2A\)

Chọn B

Câu 24.

Hai điện trở cùng mắc song song vào một mạch\(\Rightarrow U_1=U_2=U_m\) \(\Rightarrow P_1\cdot R_1=P_2\cdot R_2\)

\(\Rightarrow P_2=\dfrac{P_1\cdot R_1}{R_2}=\dfrac{5\cdot10}{20}=2,5W\)

Chọn C.

Câu 25.

\(\left(R_1ntR_2\right)//R_3\)

\(R_{tđ}=\dfrac{\left(R_1+R_2\right)\cdot R_3}{R_1+R_2+R_3}=\dfrac{\left(2+4\right)\cdot6}{2+4+6}=3\Omega\)

\(I_A=I_m=\dfrac{U_V}{R_{tđ}}=\dfrac{6}{3}=2A\)

Chọn B.

program vd1;

uses crt;

begin

clrscr;

writeln('chao cac ban toi la minh');

readln;

end.

a: vì \(8>6\\ \Rightarrow\dfrac{8}{9}>\dfrac{6}{9}\)

b: \(41>7\\ \Rightarrow\dfrac{41}{12}>\dfrac{7}{12}\)

16) ĐKXĐ: \(x\ge-\dfrac{1}{4}\)

\(\sqrt{4x+1}-\sqrt{3x+4}=1\\ \Leftrightarrow\sqrt{4x+1}=1+\sqrt{3x+4}\\ \Rightarrow4x+1=1+3x+4+2\sqrt{3x+4}\\ \Leftrightarrow x-4=2\sqrt{3x+4}\\ \Leftrightarrow x^2-8x+16=12x+16\\ \Leftrightarrow x^2-20x=0\\ \Leftrightarrow x\left(x-20\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=20\end{matrix}\right.\)

Thử lại \(x=0\) không thỏa mãn \(\Rightarrow x=20\)

\(16,ĐK:x\ge-\dfrac{1}{4}\\ PT\Leftrightarrow\sqrt{4x+1}=\sqrt{3x+4}+1\\ \Leftrightarrow4x+1=3x+4+1+2\sqrt{3x+4}\\ \Leftrightarrow x-4=2\sqrt{3x+4}\\ \Leftrightarrow x^2-8x+16=12x+16\\ \Leftrightarrow x^2-20x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=20\left(tm\right)\end{matrix}\right.\)

\(17,ĐK:x\ge2\\ PT\Leftrightarrow2x+2\sqrt{\left(x+2\sqrt{2x-4}\right)\left(x-2\sqrt{2x-4}\right)}=8\\ \Leftrightarrow4-x=\sqrt{x^2-4\left(2x-4\right)}\\ \Leftrightarrow\sqrt{x^2-8x+16}=4-x\\ \Leftrightarrow\left|x-4\right|=4-x\\ \Leftrightarrow\left[{}\begin{matrix}x-4=4-x\left(x\ge4\right)\\4-x=4-x\left(2\le x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(1,\Delta=b^2-4ac=5^2-4.2=17>0\)

=> Pt có 2n pb

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-5+\sqrt{17}}{4}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-\sqrt{17}}{4}\)

2,\(\Delta=b^2-4ac=\left(-1\right)^2-4.5.\left(-2\right)=41>0\)

=> Pt có 2n pb

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+\sqrt{41}}{10}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-\sqrt{41}}{10}\)

3,\(\Delta=b^2-4ac=\left(-1\right)^2-4.9=-35< 0\)

=> Pt vô nghiệm