Bài 3:

a. \(R=p\dfrac{l}{S}=1,1.10^{-6}\dfrac{30}{0,2.10^{-6}}=165\Omega\)

b. \(Q=UIt=220\left(\dfrac{220}{165}\right).15.60=254000\left(J\right)\)

Bài 2:

a. \(R=\dfrac{U^2}{P}=\dfrac{220^2}{1000}=48,4\Omega\)

b. \(Q=UIt=220\left(\dfrac{220}{48,4}\right).4.3600=14400000\left(J\right)\)

c. \(Q'=Q.40=14400000.40=576000000\left(J\right)=120000\)kWh

\(\Rightarrow T=Q'.2100=120000.2100=252000000\left(dong\right)\)

Bài 1:

a. \(Q=UIt=220.2,5.1=550\left(J\right)\)

b. \(\left\{{}\begin{matrix}Q_{thu}=mc\left(t-t_1\right)=2.4200\left(100-20\right)=672000\left(J\right)\\Q_{toa}=UIt=220.2,5.15.60=495000\left(J\right)\end{matrix}\right.\)

\(\Rightarrow H=\dfrac{Q_{toa}}{Q_{thu}}100\%=\dfrac{495000}{672000}100\%\approx73,6\%\)

11)11) 3x(x-5)²-(x+2)³+2(x-1)³-(2x+1)(4x²-2x+1)=3x(x²-10x+25)-(x³+6x²+12x+8)+2(x³-3x²+3x-1)-(8x³+1)=3x³-30x²+75x-x³-6x²-12x-8+2x³-6x²+6x-2-8x³-1=-4x³-42x²+63x-11

nhìn khó thế

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

\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)

\(\Leftrightarrow3-x=0\)

hay x=3

BÀI 3:

bài 4:

Là mấy bài này em làm được bài nào chưa?

\(a,4x^2-6x=2x\left(2x-3\right)\\ b,9x^4y^3+3x^2y^4=3x^2y^3\left(2x^2+y\right)\\ c,x^3-2x^2+5x=x\left(x^2-2x+5\right)\\ d,3x\left(x-1\right)+5\left(x-1\right)=\left(3x+5\right)\left(x-1\right)\\ e,2x^2\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(2x^2+4\right)=2\left(x+1\right)\left(x^2+2\right)\\ f,2x^2y-4xy^2+6xy=2xy\left(x-y+3\right)\\ g,4x^3+4x^2+4x=4x\left(x^2+x+1\right)\\ h,x^3+x^2-3x-27=x^3-3x^2+4x^2-12x+9x-27=x^2\left(x-3\right)+4x\left(x-3\right)+9\left(x-3\right)=\left(x^2+4x+9\right)\left(x-3\right)\\ i,4x^2-12x+9=\left(2x-3\right)^2\\ k,8x^3-27=\left(2x\right)^3-3^3=\left(2x-3\right)\left(4x^2+6x+9\right)\\ l,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

Tick nha 😘

Đây nè bạn đã cố gắng ko làm tắt rồi nhé bạn

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)

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

d: \(3x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(3x+5\right)\)

e: \(2x^2\left(x+1\right)+4\left(x+1\right)=2\left(x+1\right)\left(x^2+2\right)\)

f: \(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)

nhấn F11

bạn xài laptop hay máy tính bàn

hay là máy tinha của bn bị vào chế độ toàn màn hình rồi, nếu thế thì bn thử ấn f11 xem sao

1) \(x^3-8x+7=\left(x-1\right)\left(x^2+x-7\right)\)

2) \(x^3+8x^2-9=\left(x-1\right)\left(x^2+9x+9\right)\)

3) \(3x^3-4x+1=\left(x-1\right)\left(3x^2+3x-1\right)\)

4) \(x^4-3x^2+3x-1=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)

5) \(x^4-5x^2+4=\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)\)

1: Ta có: \(x^3-8x+7\)

\(=x^3-x-7x+7\)

\(=x\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x-7\right)\)

2: Ta có: \(x^3+8x^2-9\)

\(=x^3-x^2+9x^2-9\)

\(=x^2\left(x-1\right)+9\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^2+9x+9\right)\)

3: Ta có: \(3x^3-4x+1\)

\(=3x^3-3x-x+1\)

\(=3x\left(x-1\right)\left(x+1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(3x^2+3x-1\right)\)

4: Ta có: \(x^4-3x^2+3x-1\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-3x\cdot\left(x-1\right)\)

\(=\left(x-1\right)\cdot\left(x^3+x+x^2+1-3x\right)\)

\(=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)

5: Ta có: \(x^4-5x^2+4\)

\(=x^4-x^2-4x^2+4\)

\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2-4\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)

a: Ta có: \(2x+3=x+1\)

\(\Leftrightarrow2x-x=1-3\)

hay x=-2

b: Ta có: \(2x\left(2x-1\right)-\left(2x+3\right)^2=5\)

\(\Leftrightarrow4x^2-2x-4x^2-12x-9=5\)

\(\Leftrightarrow-14x=14\)

hay x=-1

c: Ta có: \(4x^2-25\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(2x-5x-10\right)\left(2x+5x+10\right)=0\)

\(\Leftrightarrow\left(-3x-10\right)\left(7x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=-\dfrac{10}{7}\end{matrix}\right.\)

d: Ta có: \(2x^2+7x+5=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e: Ta có: \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

f: Ta có: \(\dfrac{1}{9}x^3-x=0\)

\(\Leftrightarrow x\left(\dfrac{1}{9}x^2-1\right)=0\)

\(\Leftrightarrow x\left(\dfrac{1}{3}x-1\right)\left(\dfrac{1}{3}x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

g: Ta có: \(x^3+3x^2+3x=7\)

\(\Leftrightarrow\left(x+1\right)^3=8\)

\(\Leftrightarrow x+1=2\)

hay x=1

a, x = -2

b, x = -1

c, x = -10/3 hoặc -10/7

kb nha

k nha

Trò chuyện vs mk nhé bn

nếu muốn hết buồn bạn hãy xem hài cho hết buồn