giúp mình với, mình đang cần gấp, mình cảm ơnnn
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NN
1
15 tháng 12 2022
a: x=3/4-1/3=9/12-4/12=5/12
b: 1/3x=2/3-5/9=6/9-5/9=1/9
=>x=1/3
c: =>3x=6/5-3/4-7/4=6/5-5/2=12/10-25/10=-13/10
=>x=-13/30
d: =>x+2/5-2/3=5/3
=>x=5/3+2/3-3/5=7/3-3/5=35/15-9/15=26/15
e: =>1/4:x=2/5-3/4=8/20-15/20=-7/20
=>x=-1/4:7/20=-1/4*20/7=-20/28=-5/7
f: =>1/4x-3/4=1/2-25/4=2/4-25/4=-23/4
=>1/4x=-20/4
=>x=-20
DN
Giúp mình với ạ, mình cảm ơnnn
1
19 tháng 12 2022
Bài 6.
a)Công suất ấm: \(P=\dfrac{A}{t}=\dfrac{900\cdot1000}{10\cdot60}=1500W\)
Dòng điện qua ấm: \(I=\dfrac{P}{U}=\dfrac{1500}{220}=\dfrac{75}{11}A\)
Điện trở dây nung: \(R=\dfrac{U}{I}=\dfrac{220}{\dfrac{75}{11}}=\dfrac{484}{15}\Omega\)
b)Điện năng tiêu thụ trong 1 tháng (30 ngày):
\(T=900\cdot1000\cdot30\cdot3600=9,72\cdot10^{10}J=27000kWh\)
Tiền điện phải trả: \(T=27000\cdot1500=40500\left(k.đồng\right)\)
c)Công suất tiêu thụ thực:
\(P=UI=\dfrac{U^2}{R}=\dfrac{110^2}{\dfrac{484}{15}}=375W\)
19 tháng 12 2022
Bài 7.
CTM: \(\left(Đ_1ntR_b\right)//Đ_2\)
\(R_1=\dfrac{U_{Đ1}^2}{P_{Đ1}}=\dfrac{10^2}{2}=50\Omega;I_{Đ1đm}=\dfrac{P_{Đ1}}{U_{Đ1}}=\dfrac{2}{10}=0,2A\)
\(R_2=\dfrac{U^2_{Đ2}}{P_{Đ2}}=\dfrac{12^2}{3}=48\Omega;I_{Đ2đm}=\dfrac{P_{Đ2}}{U_{Đ2}}=\dfrac{3}{12}=0,25A\)
Để đèn 1 sáng bình thường \(\Rightarrow I_b=I_{Đ1đm}=0,2A\)
\(R_{Đ1+b}=\dfrac{12}{0,2}=60\Omega\)
\(R_b=60-R_{Đ1}=60-50=10\Omega\)
giúp e với đang cần thực sự gấp ạ! e cảm ơnnn
4 tháng 9 2021
a: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)
\(\Leftrightarrow9x^3+6x^2+27x+28-9x^3-6x^2-x=54\)
hay x=1
b: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
hay x=-1
H9
HT.Phong (9A5)
CTVHS
28 tháng 10 2023
Bài 2:
\(A=2+2^2+2^3+...+2^{60}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(A=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)
\(A=3\cdot\left(2+2^3+...+2^{59}\right)\) ⋮ 3
Vậy: A ⋮ 3
_____________
\(A=2+2^2+...+2^{60}\)
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(A=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)
\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)
\(A=7\cdot\left(2+2^4+....+2^{58}\right)\) ⋮ 7
Vậy: A ⋮ 7
___________________
\(A=2+2^2+...+2^{60}\)
\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)
\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)
\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)
\(A=5\cdot\left(2+2^2+...+2^{58}\right)\) ⋮ 5
Vậy: A ⋮ 5
TT
1
2 tháng 6 2021
\(A=\left(\frac{x}{x+2}+\frac{x^3}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{x^2-2x+4}{4-x^2}\right):\frac{4}{x+2}\)
\(A=\left(\frac{x}{x+2}+\frac{x^3}{x+2\left(4-x^2\right)}\right):\frac{4}{x+2}\)
\(A=\left(\frac{4x-x^3+x^3}{x+2\left(4-x\right)}\right):\frac{4}{x+2}\)
\(A=\frac{4x}{x+2\left(4-x\right)}.\frac{x+2}{4}\)
\(A=\frac{x}{4-x}\)
\(b,\frac{x}{4-x}>0\)
xét 2 trường hợp x>0 đồng thời 4-x>0 (điều kiện x\(\ne\)4) và x<0 ,4-x<0
\(TH1:0< x< \text{4}\)
\(TH2:\)ko có giá trị x
\(c,Ax=\frac{x}{4-x}x\)=\(\frac{x^2}{4-x}\)
\(\frac{x^2-16+16}{4-x}\)
\(\frac{\left(x-4\right)\left(x+4\right)+16}{4-x}\)
\(-\left(x+4\right)+\frac{16}{4-x}\)
để AX nguyên thì \(16⋮4-x\)
lập bảng ra tìm đc x = 0,2,-4,-12,5,6,8,12,20
3 C
4 B
5 A
6 B
7 A
8 A
9 C
10 D
11 B
12 B
13 A
14 C
15 B
16 A
17 D
18 D
19 C
20 A
21 B
22 B
23 C
24 D
25 B
26 D
27 A
28 D
29 C
30 B
31 C
32 A
33 A
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