Bài 1:

\(54\left(\dfrac{km}{h}\right)=15\left(\dfrac{m}{s}\right);9\left(\dfrac{m}{s}\right)=32,4\left(\dfrac{km}{h}\right)\)

Baì 2:

\(t'=s':v'=5:\left(5.3,6\right)=\dfrac{5}{18}h\)

\(\Rightarrow v_{tb}=\dfrac{s'+s''}{t'+t''}=\dfrac{5+3,8}{\dfrac{5}{18}+\left(\dfrac{15}{60}\right)}\simeq16,67\left(\dfrac{km}{h}\right)\)

Câu 2: 

\(\Leftrightarrow\left(x+2\right)\left(10x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{3}{10}\end{matrix}\right.\)

Áp dụng bất đẳng thức Cosi ta có :

\(x^4+1\ge2x^2;x^2+1\ge\left|x\right|\Rightarrow x^4+3\ge4\left|x\right|\)

Tương tự : \(y^4+3\ge4\left|y\right|\)

\(\Rightarrow x^4+y^4+6\ge4\left(\left|x\right|+\left|y\right|\right)\left(1\right)\)

Từ (1) suy ra \(x^4+y^4+6\ge4\left(x-y\right)\Rightarrow P\le\dfrac{1}{4}\)

Dấu = xảy ra \(x=1;y=-1\)

Từ (1) suy ra \(x^4+y^4+6\ge4\left(y-x\right)\Rightarrow P\ge-\dfrac{1}{4}\)

Dấu = xảy ra \(x=-1;y=1\)

Bài 1.

a)Điện trở tương đương: \(R_m=R_1+R_2=12+8=20\Omega\)

b)\(I_A=I_1=I_2=\dfrac{U_{AB}}{R_m}=\dfrac{18}{20}=0,9A\)

c)\(U_1=I_1\cdot R_1=0,9\cdot12=10,8V\)

  \(U_2=I_2\cdot R_2=0,9\cdot8=7,2V\)

d)\(R_Đ=\dfrac{U_Đ^2}{P_Đ}=\dfrac{12^2}{6}=6\Omega\)

   \(\Rightarrow R_m=R_1+R_Đ=12+6=18\Omega\)

   \(I_m=\dfrac{U}{R}=\dfrac{18}{18}=1A\)

   \(I_{Đđm}=\dfrac{P_Đ}{U_Đ}=\dfrac{6}{12}=0,5A< I_m=1A\)

   Vậy đèn sáng yếu hơn bình thường.

Bài 2:

a. \(R=\dfrac{R1.R2}{R1+R2}=\dfrac{20.30}{20+30}=12\Omega\)

\(U=U1=U2=IR=12.2=24V\left(R1\backslash\backslash\mathbb{R}2\right)\)

b. \(\left\{{}\begin{matrix}I1=U1:R1=24:20=1,2A\\I2=U2:R2=24:30=0,8A\end{matrix}\right.\)

c. \(I=I12=I3=0,5A\left(R12ntR3\right)\)

\(U3=U-U12=24-\left(0,5.12\right)=18V\)

d. \(P=UI'=24.0,5=12\)W

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