giúp câu b ặ;-;
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`4` If the weather is nice, we will go swimming with my friends.
1: Thay x=25 vào B, ta được:
\(B=\dfrac{1}{5-2}=\dfrac{1}{3}\)
2; P=A:B
\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{1}\)
\(=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
\(P=A:B=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\right):\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{x+2-2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-2\right)\\ =\dfrac{x+2-2x+4\sqrt{x}+x-1}{\sqrt{x}+1}\\ =\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
\(R_Đ=\dfrac{U^2_Đ}{P_Đ}=\dfrac{12^2}{6}=24\Omega\)
\(R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{20\cdot20}{20+20}=10\Omega\)
\(R_{tđ}=R_Đ+R_{12}=24+10=34\Omega\)
\(I_A=I_m=\dfrac{U_m}{R_{tđ}}=\dfrac{15}{34}=0,44A\)
\(\dfrac{x+1}{3}=\dfrac{y+2}{4}=\dfrac{z+3}{5}\\ =\dfrac{x+1+y+2+z+3}{3+4+5}=\dfrac{24}{12}=2\)
Từ:
\(\dfrac{x+1}{3}=2\\ =>x+1=6\\ < =>x=5\)
\(\dfrac{y+2}{4}=2\\ =>y+2=8\\ < =>y=6\)
\(z=18-x-y=18-5-6=7\)