If f (2x + 1) = the square of 2x - 3x + 5, the solution of F (X-2) can be obtained by substitution

If f (2x + 1) = the square of 2x - 3x + 5, the solution of F (X-2) can be obtained by substitution

Let A-2 = 2x + 1
x=(a-3)/2
So f (A-2) = 2 [(A-3) / 2] &# 178; - 3 (A-3) / 2 + 5
=(a²-6a+9-3a+9+10)/2
=(a²-9a+28)/2
So f (X-2) = (X & # 178; - 9x + 28) / 2
How to divide the solution 1 by (the square of X + 2x + 3) in the first substitution method of indefinite integral?
∫ dx/(x² + 2x + 3)
= ∫ dx/[(x + 1)² + 2]
Let z = (x + 1) / √ 2, DZ = (1 / √ 2) DX
= ∫ √2•dz/(2z² + 2)
= √2/2•∫ dz/(z² + 1)
= (1/√2)arctan(z) + C
= (1/√2)arctan[(x + 1)/√2] + C
The quadratic function collocation method is used to solve y = x square - x + 2, and the substitution method is used to solve y = 2x + ≤ √ 2-x
Solution of y = xsquare-x / xsquare-x + 1 by separation constant method
The formula is not well expressed and can't be solved
Solving x ^ 2 / (x-1) ^ 2-2x / X-1 = 3 by substitution method
Solving x ^ 2 / (x-1) ^ 2-2x / X-1 = 3 by substitution method
Let y = x / (x-1)
The original equation is reduced to y ^ 2-2y-3 = 0
The solution is y = 3, y = - 1
When x / (x-1) = 3, x = 3 / 2
When x / (x-1) = - 1, x = 1 / 2
The test shows that X1 = 3 / 2, X2 = 1 / 2 are the solutions of the original equation
The solution of the original equation is X1 = 3 / 2, X2 = 1 / 2
Let X / (x-1) = a
a^2-2a-3=0
(a-3)(a+1)=0
A = 3 or - 1
If x / X-1 = 3, then x = 3 / 2
If x / X-1 = - 1, x = 1 / 2
x/x-1=y,
y^2-2y=3,
y=3,-1.
x=2/3,1/2
For example, internal energy, mechanical energy, electrical energy, chemical energy and light energy are transformed into each other
Inside the machine, the water rolls and the flute sounds
Internal friction generates heat
Electromagnetic induction of generator
Electric fan
Electrochemical electrolysis
Chemical battery
Chemical --- light. Internal combustion
Photo electric solar cell
Photochemical photosynthesis
What is the rated voltage of an incandescent lamp marked with 220v100w when it is connected to a circuit with 110V voltage?
What is the rated power? What is the actual voltage? What is the actual power? Its brightness will be higher than that under the rated voltage___ .
The rated voltage is 220 V and the rated power is 100 W,
The resistance of the lamp is r = u ^ 2 / P = 220 ^ 2 / 100 = 484 Ω
The actual power is p = u ^ 2 / r = 110 ^ 2 / 484 = 25W,
The actual voltage is 110V,
Since the actual power is less than the rated power,
So its brightness is lower than that under rated voltage
Actual voltage 110V, rated power 100W
R=U^2/P=220×220/100=484Ω
Actual power p '= u' ^ 2 / r = 110 × 110 / 484 = 25W
Its brightness will be darker than that at rated voltage
What is the rated voltage of an incandescent lamp marked with 220v100w when it is connected to a circuit with 110V voltage?
The question is simple, ha ha.
Rated power of incandescent lamp = 100W
The actual voltage is 110V
The actual power is 25W
Its brightness is darker than that under rated voltage.
Four mixed operations of 500 integers
Do not apply the problem, must be simple
Don't answer me
1) 86+49+114= 2) 240+(39-40)=
3) 255+(352+145+48)= 4) (345+377)+(55+23)=
5) 9+(80+191)= 6) (268+314+132)+86=
7) 5190÷15= 8) 495+(278+5)+222=
9) 174×36×25= 10) 399-199=
11) 48+(164+152)+36= 12) 133-(28+29)-43=
13) 1650÷25= 14) 260×8-8-8×59=
15) 996+500= 16) 6975÷25=
17) 196-95= 18) 328-(163-72)=
19) 199+(84-99)= 20) 885-1-201-298=
21) 460-35-3-262= 22) (98+59+2)+41=
23) 736×12-12-12×335= 24) 116+(112+184)=
25) 150×258+142×150= 26) 31×24×25=
27) 9000÷25= 28) 502-287-54-159=
29) 307+(92+93)= 30) 80×125=
31) 102×15= 32) 30+(63+70)+37=
33) 27+(73+73)+27= 34) 86+(98+14+2)=
35) 544-272-28= 36) 18000÷150÷4=
37) 103×69= 38) 25×64×125=
39) 343-188-12= 40) 509×11-11-11×8=
41) 79×24×25= 42) (145+25)+(155+275)=
43) (447+423)+(53+77)= 44) 46+15+54=
45) 589-109-(6+185)= 46) 8×125=
47) 20×25= 48) 89×245+155×89=
49) 92+(79+8+21)= 50) 222+15+78=
51) 96×125= 52) 30600÷25÷4=
53) 5996+3004= 54) 6015-(518+699)-2783=
55) 4003×2426= 56) 2467×70-70-70×466=
57) 84×25= 58) 4001-2002=
59) 1616×506+2494×1616= 60) 4×17+4+1982×4=
61) 799×660+340×799= 62) 3991×36×25=
63) 6076-875-(805+3320)= 64) 6056-679-40-4281=
65) 4134+(2819+866)+2181= 66) 5898-(2065-102)=
67) 3297×1273+2727×3297= 68) 1312+(153+688+1847)=
69) 2315-793-114-1093= 70) 3940+(1739-1940)=
71) 1455+(1768+1545)+1232= 72) 975+(1007+2025)=
73) 24×1951+24+48×24= 74) 30425÷25=
75) 1376+(1961+624+39)= 76) (686+1872+2314)+1128=
77) 2922+(260-922)= 78) 113600÷100÷4=
79) 2002×658= 80) 1428+(958+2572)=
81) 2001×786= 82) 190×760+190+3239×190=
83) 2976×1145+2855×2976= 84) 88×25=
85) 8122-(3084-1878)= 86) 879+(1295+2121)=
87) 3998+2001= 38) 2595×178-178-178×594=
89) 4467-2024-976= 90) 1319×1339+1661×1319=
91) 997×885= 92) 453×8×125=
93) 4928-(871+1928)= 94) 997×917=
95) 1526+(938-526)= 96) 803×12×25=
97) 114000÷1200= 98) 6933×332-332-332×2932=
99) 16×25= 100) 25×224×125=
101)9/22 + 1/11 ÷ 1/2
102)5/3 × 11/5 + 4/3
103)45 × 2/3 + 1/3 × 15
104) 7/19 + 12/19 × 5/6
106) 8/7 × 21/16 + 1/2
107) 101 × 1/5 – 1/5 × 21
108)50＋160÷40 （58+370）÷（64-45）
109)120-144÷18+35
110)347+45×2-4160÷52
111)（58+37）÷（64-9×5）
112)95÷（64-45）
113)178-145÷5×6+42 420+580-64×21÷28
114)812-700÷（9+31×11） （136+64）×（65-345÷23）
115)85+14×（14+208÷26）
116)（284+16）×（512-8208÷18）
117)120-36×4÷18+35
118)（58+37）÷（64-9×5）
119)(6.8-6.8×0.55)÷8.5
120)0.12× 4.8÷0.12×4.8
121)（3.2×1.5+2.5）÷1.6 （2）3.2×（1.5+2.5）÷1.6
121)6-1.6÷4= 5.38+7.85-5.37=
122)7.2÷0.8-1.2×5= 6-1.19×3-0.43=
123)6.5×（4.8-1.2×4）= 0.68×1.9+0.32×1.9
124)10.15-10.75×0.4-5.7
125)5.8×（3.87-0.13）+4.2×3.74
126)32.52-（6+9.728÷3.2）×2.5
127)[（7.1-5.6）×0.9-1.15] ÷2.5
128)5.4÷[2.6×（3.7-2.9）+0.62]
129）0.9+0.1÷0.1 0.3×0.3×0.3
130）0.5÷0.5÷0.5 0.8-0.8×0.5
131）0.8÷0.8×0.5 2.7+2.3÷0.2
132）5.4÷1.8-1.8 11.2-1.93+0.8
133）0.38×2.9+0.38 0.5-0.5×0.5
134）0.8÷0.8×0.5 2.7+2.3÷0.2
135）5.4÷1.8-0.8 11.2-1.93+8.07
136）1-1÷4 0.65×102
137）9.87-（5.87+2.9）
138）（0.25+0.45）×0.4
139）（0.36+1.29）÷3 0.008+0.992×2.5×40
140）4.84+0.3×15÷0.2+77.5 0.15×（3.79-1.9）+1.11×0.15
141）0.05×[30-(18.4+27.83÷4.6)] (6.8-6.8×0.55)÷8.5
142）0.12× 4.8÷0.12×4.8 1.6-1.6÷4
143）5.38+7.85-5.37 7.2÷0.8-1.2×5
144）6-1.19×3-0.43 6.5×（4.8-1.2×4）
145）0.68×1.9+0.32×1.9 10.15-10.75×0.4-5.7
147）146）5.8×（3.87-0.13）+4.2×3.74 32.52-（6+9.728÷3.2）×2.5
148）[（7.1-5.6）×0.9-1.15] ÷2.5 5.4÷[2.6×（3.7-2.9）+0.62]
149）5.47+12.81+3.53+7.19 0.83×12.5×8
2.9×102 3.8×6.9+3.8×2.1+3.8
150）109+(72+91)-93×24×125=
There is no more
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Xiao Ming and Xiao Gang collect stamps. The number of stamps collected by Xiao Ming is three times that of Xiao Gang. Xiao Gang collects 64 fewer stamps than Xiao Ming. How many stamps have Xiao Gang and Xiao Ming collected?
Specific list formula, and make a detailed explanation
Suppose Xiao Ming collected x stamps, while Xiao Gang collected 3x stamps
3x - x = 64
x = 32
3x = 96
A: Xiao Ming collected 32 stamps and Xiao Gang 96 stamps
Please give me an example of converting chemical energy into electrical energy in our daily life
Battery. Zinc as negative electrode, carbon as positive electrode
How many acres of land is 503 square meters
7545 Mu
It should be less than one mu