giúp e với ạ e cảm ơn nhiềuuu
các vị giúp e giải bài này với ạ. e cảm ơn rất nhiềuuu :3
Giúp với ạ, cảm ơn nhiềuuu
\(\left\{{}\begin{matrix}P=U.I\Rightarrow I=\dfrac{P}{U}=\dfrac{100}{220}=\dfrac{5}{11}\left(A\right)\\R=\dfrac{U}{I}=\dfrac{220}{\dfrac{5}{11}}=484\left(\Omega\right)\end{matrix}\right.\)
\(R_Đ=\dfrac{U_Đ^2}{P_Đ}=\dfrac{220^2}{100}=484\Omega\)
a)Cường độ dòng điện:
\(I_m=\dfrac{U_m}{R}=\dfrac{220}{484}=\dfrac{5}{11}A\)
b)Cường độ dòng điện khi đặt 1 hđt 200V vào hai đầu đèn:
\(I'=\dfrac{200}{484}=\dfrac{50}{121}A\)
Điện năng đèn tiêu thụ trong 10' :
\(A=UIt=200\cdot\dfrac{50}{121}\cdot10\cdot60=49586,8J\)
Giúp mình với ạ:== cảm ơn mn nhiềuuu
II
1 C
2 B
3 C
4 D
5 A
6 A
7 D
8 C
III
1 is getting
2 is - is visited
3 selling
4 were you doing - phoned
IV
1 This road isn't used very often
2 A lot of trees will be grown in the park
3 The living-room is being painted light blue
4 Many different paper products have been developed
5 We were asked many difficult questions by the examier
V
1 It's is coffee "black" without cream or sugar
2 They are China, Japan and other Oriental countries
3 They drink tea with sugar
4 Yes, it is
VI
1 drive => driving
2 That => the
3 meets => has met
4 getting => to get
Đề 2
Bài 1
1C
2D
3D
4C
Bài 2
1 C
2 A
3 C
4 B
5D
6 D
7 C
8 D
9 B
10 A
Bài 3
1 is going to rain
2 haven't invited
3 is getting
4 saw
Giúp mình phần b với ạ, mình cảm ơn nhiềuuu
Thay m=2 vào HPT ta có:
\(\left\{{}\begin{matrix}2x+y=1\\x+2y=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}4x+2y=2\\x+2y=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}4x+2y=2\\3x=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}mx+y=1\\x+my=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=1-mx\\x+m\left(1-mx\right)=1\left(1\right)\end{matrix}\right.\)
(1) ⇔x+m-m2x=1
⇔x(1-m2)=1-m (2)
TH1: 1-m2 = 0
⇔m = +- 1
Thay m=1 vào (2) ta có: 0x=0 (Luôn đúng) ⇒m=1 (chọn)
Thay m=-1 vào (2) ta có: 0x=2 (Vô lí) ⇒m=-1 (loại)
TH2: 1-m2 ≠0
⇔m≠ +-1
⇒HPT có nghiệm duy nhất:
x= \(\dfrac{1-m}{1-m^2}\)
⇒y= \(1-m.\dfrac{1-m}{1-m^2}\)
⇔y=\(\dfrac{1-m}{1-m^2}\)
Dễ thấy x=y nên:
\(\dfrac{1-m}{1-m^2}>0\)
⇔1-m>0
⇔m<1
Vậy m <1 thì Thỏa mãn yêu cầu đề bài.
a) Thay m=2 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}2x+y=1\\x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=1\\2x+4y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=-1\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\2x=1-y=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: Khi m=2 thì hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Làm giúp mik 2 bài này với, 2 bài tik 2 lần ạ , cảm ơn nhiềuuu
Giúp tui vs ạ:== cảm ơn mn nhiềuuu
1 We are having a wonderful time at the moment in Sa Pa
2 It is necessary to cool the burn imediately
1.We are having a wonderful time at the moment in Sa Pa
2.It is necessary to cool the burn imediately
Giúp minh vs ạ !! Cảm ơn mn nhiềuuu
V
1 will be watching
2 will make
3 is going to rain
4 will become - won't get
5 will change
6 to call
7 will be lying
8 had come - had been
9 arrived - had been waiting
10 are not
11 pollution
12 were destroyed
13 would do
14 had left - came
15 didn't go - stayed
16 Has - learnt , has learnt
17 Did - wear , didn't wear
18 Does - read , watch
19 would save
20 will have
21 will be being painted
22 smoked
23 came - had left
24 had finished
25 was built
VI
1
a)
1 T
2 F
3 T
4 T
5 T
6 F
b)
1 It is a device which is used to send multiple messages over a single wire
2 It is a device which is used to draw the shape of the sound waves
3 It happened in 1876, while bell was at one end of ........
Mng ơiii, chỉ giúp em câu 6 với ạ, câu hỏi là nằm về hai phía trục hoành ạ. Em cảm ơn nhiềuuu
Để hàm bậc 3 có 2 cực trị nằm về 2 phía trục hoành
\(\Leftrightarrow y=0\) có 3 nghiệm pb
\(\Leftrightarrow x^3-\left(2m+1\right)x^2+\left(m+1\right)x+m-1=0\) có 3 nghiệm pb
\(\Leftrightarrow\left(x-1\right)\left(x^2-2mx-m+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2-2mx-m+1=0\left(1\right)\end{matrix}\right.\)
Bài toán thỏa mãn khi (1) có 2 nghiệm pb khác 1
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=1-2m-m+1\ne0\\\Delta'=m^2+m-1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne\dfrac{2}{3}\\\left[{}\begin{matrix}m< \dfrac{-1-\sqrt{5}}{2}\\m>\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
Có 19 số tự nhiên nhỏ hơn 20 thỏa mãn
Làm được tới đâu thì làm giúp em ạ, cảm ơn nhiềuuu!!
a) 1 = 12/12
b) -5 = -60/12
c) -3/4 = -9/12
d) 0 = 0/12