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:

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

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

Nhường e cái lý 8 với ạ :((

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

1 more difficult

2 warmer

3 the most intelligent

4 the hottest

5 cheaper than

6 luckiest

7 more comfortable than

8 the most boring

9 the luckiest

10 simper

Thank nha

Bài 1:

\((n+1)^n-1=n[(n+1)^{n-1}+(n+1)^{n-2}+....+(n+1)+1]\)

Giờ ta chỉ cần cmr \((n+1)^{n-1}+(n+1)^{n-2}+...+(n+1)+1\vdots n\)

Thật vậy:

\((n+1)^{n-1}+(n+2)^{n-2}+...+(n+1)+1\equiv 1^{n-1}+1^{n-2}+...+1^1+1=n\equiv 0\pmod n\)

Do đó ta có đpcm.

Bài 2 em xem lại. Số $2^{n(2^n-1)}$ chỉ toàn ước có dạng $2^k$ với $k=0,1,..., n(2^n-1)$ trong khi đó $(2^n-1)^2$ là số lẻ.

\(3.\)

\(a,\)

\(\left(2x-3\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow4x^2-12x+9-x^2-10x-25=0\)

\(\Leftrightarrow3x^2-22x-16=0\)

\(\Leftrightarrow3.\left(x-8\right)\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3=0\left(\text{vô lí}\right)\\x-8=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(S=\left\{8;-\dfrac{2}{3}\right\}\)

\(b,\)

\(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{1;2\right\}\)

\(4.\)

\(a,\)

\(16x^3y+\dfrac{1}{4}yz^3\)

\(=\dfrac{1}{4}y\left(64x^3+z^3\right)\)

\(=\dfrac{1}{4}y\left(4x+z\right)\left(16x^2-4xz+z^2\right)\)

\(b,\)

\(x^{m+4}-x^{m+3}-x-1\)

\(=x^m.x^4-x^m.x^3-x-1\)

\(=x^m.\left(x^4-x^3\right)-x-1\)

\(=x^m.x^3.\left(x+1\right)-\left(x+1\right)\)

\(=\left(x^{m+3}-1\right)\left(x+1\right)\)

3:

a: =>(2x-3-x-5)(2x-3+x+5)=0

=>(x-8)(3x+2)=0

=>x=8 hoặc x=-2/3

b: =>x^3-x^2-4(x-1)^2=0

=>x^2(x-1)-4(x-1)^2=0

=>(x-1)(x^2-4x+4)=0

=>x=1 hoặc x=2