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

24 thì sao ah,cảm ơn nhìu nha<3

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 \(\Rightarrow x=20\)

câu 17 nữa ah:> tks

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

=> 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

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

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