If a root of the equation 2x & # 178; + (T + 3) x-4 = 0 about X is a, and A-2 / a = 6, find the value of T

If a root of the equation 2x & # 178; + (T + 3) x-4 = 0 about X is a, and A-2 / a = 6, find the value of T

A-2 / a = 6, the square of A-2 = 6A (1) is obtained by removing the denominator;
Because a is the root of the equation: the square of 2A + (T + 3) A-4 = 0,;
Multiply (1) by 2 to get the square of 2A - 4 = 12a (2),
Comparing (3) with (2), we find that - (T + 3) a = 12a;
So - (T + 3) = 12;
t=-15
Substituting a into the equation, t = (4-2a ^ 2-3a) / A is obtained
From (A-2) / a = 6, a = - 2 / 5 can be obtained by substituting (A-2) / a = 6?
Is the light of electric lamp converted from electric energy or from internal energy converted from electric energy?
If it is converted from internal energy, it should not be a pure resistance appliance. But it is said in the book that it is a pure resistance appliance? What should I do in the exam?
The light-emitting of electric lamp is the reason that the electric energy is converted into internal energy. The filament is a pure resistance, and the current is converted into internal energy through the filament. The temperature of the filament rises. When the filament temperature reaches a certain level, part of its internal energy is converted into light energy and transmitted outward in the form of radiation
The working current of fluorescent lamp is 150mA, closing () a, closing () UA
0.15A,150000uA
Operation problems of integral
1. Given a ^ 2 + B ^ 2 + 2a-4b + 5 = 0, then the value of 2A ^ 2 + 4ab-3 is ();
2. Given that the rational number x satisfies 4x ^ 2-4x + 1 = 0, the value of the algebraic formula 2x + 1 / 2x is ()
(there should be a problem-solving process)
1.a^2+b^2+2a-4b+5=0
a^2+2a+1+b^2-4b+4=0
（a^2+2a+1）+（b^2-4b+4）=0
（a+1)^2+（b-2）^2=0
Because the sum of squares is greater than or equal to 0, the sum of two numbers greater than or equal to zero equals zero, and only two formulas are 0
So a + 1 = 0, B-2 = 0, a = - 1, B = 2
2a^2+4ab-3= 2*（-1）^2+4*(-1)*2-3=-9
2. X satisfies 4x ^ 2-4x + 1 = 0
(2x-1)^2=0
x=1/2
(2x + 1 / 2x, is the last term here 2x 1 / 2 or 1 / 2 x?)
1 / 2 of X
2x+1/2x=2*1/2+1/2*1/2=1.25
One out of two
2x+1/2x=2*1/2+1/（2*1/2）=2
Given Tana = 3, calculate (5cos ^ 2a-3sin ^ 2a) / (1 + sin ^ 2a)
A:
tana=3,sina=3cosa
Substituting Sin & # 178; a + cos & # 178; a = 1, we get: cos & # 178; a = 1 / 10
(5cos²a-3sin²a) /(1+sin²a)
=(5-3tan²a) /(1/cos²a+tan²a)
=(5-27) / (10+9)
=-22/19
This is the first time that we will be able to reach a-3sin-3d-and-#178; a-3sin & #178; a-3sin & #178; a) / / (1 + Sin & #178; a) / / (1 + sin #35;178; a) / (1 + sin ##178; 178; a) / (1 + sin \35;##35\35\\\\35;178; 178; a) = (5-3-3tan & \\\\\\\\#178; 178; 178; 178; 178; 178; 178; 178; 178; (5-3d-3d-3d-3d-3d-3d-3d-3d-3d-3d-3d-x 3 & # 178; + 1) & nbsp; = (5-27) / (18 + 1) = - 22 / 19 & nbsp;
simple form
=(5cos^2a-3sin^2a)/(sin^2a+cos^2a+sin^2a)
=(5cos^2a-3sin^2a)/(2sin^2a+cos^2a)
Divide cos ^ 2A
=(5-3tan^2a)/(2tan^2a+1)
=(5-3*3^2)/(2*3^2+1)
=(5-3*9)/(2*9+1)
=-22/19
... unfold
simple form
=(5cos^2a-3sin^2a)/(sin^2a+cos^2a+sin^2a)
=(5cos^2a-3sin^2a)/(2sin^2a+cos^2a)
Divide cos ^ 2A
=(5-3tan^2a)/(2tan^2a+1)
=(5-3*3^2)/(2*3^2+1)
=(5-3*9)/(2*9+1)
=-22/19
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I wish you progress!
If you don't understand, please ask in time. If you are satisfied, please accept_ ∩) O thank you ~ ~ put it away
(5cos^2a-3sin^2a)/(1+sin^2a)
=(5cos^2a-3sin^2a)/(sin^2a+cos^2a+sin^2a)
=(5cos ^ 2a-3sin ^ 2a) / (2Sin ^ 2A + cos ^ 2a) (numerator and denominator divided by cos ^ 2a)
=(5cos^2a/cos^2a-3sin^2a/cos^2a)/(2sin^2a/cos^2a+cos^2a/cos^2a)
=(5-3tan^2a)/(2tan^2a+1)
=(5-3*3^2)/(2*3^2+1)
=(5-27)/(18+1)
=-22/19
The efficiency of converting electric energy into light when incandescent lamp works normally
Incandescent lamp produces more heat
In incandescent lamps and energy-saving lamps, the same rated power means that the same amount of electric energy is converted into light energy and internal energy in the same time. The energy-saving lamps have less internal energy and more light energy, so the efficiency is high
Incandescent lamps should be converted twice, energy-saving lamps should only be converted once
Incandescent lamp is the conversion of electric energy into internal energy and then to light energy, which causes great loss
The energy-saving lamp is the direct conversion of electric energy into light energy, the loss is small. The other part of the conversion into internal energy is not as much as incandescent lamp
Incandescent lamp, is the conversion of electrical energy into heat energy. Part of the heat energy into radiant energy (light); a conversion
The energy-saving lamp is a kind of energy-saving lamp, which converts electric energy into short wave radiation energy (electron bombards mercury atom), and then excites phosphor by part of short wave radiation energy (254nm radiation) to make phosphor emit light
The current of rice cooker is 4.2 (fill in a mA.uA )
Fill in A. according to I = P / u = P / 220 V, the general utility power is 220 V, and the power of rice cooker is from several hundred watts to several thousand watts, so the current should be ampere level, not milliampere level, not even microampere level
What is the operation of integral
Monomials and polynomials are collectively called integers
A rational expression in an algebraic expression, which does not include division or fraction, and which has division and fraction but does not include variable in division or denominator, is called integral
Integral can be divided into definition and operation, definition can be divided into monomial and polynomial, operation can be divided into addition and subtraction and multiplication and division
Addition and subtraction include merging similar terms. Multiplication and division include basic operations, rules and formulas. Basic operations can be divided into the operation properties of power. Rules can be divided into integral and division, and formulas can be divided into multiplication formula, zero exponent power and negative integer exponent power
1、 Four operations of integral
1. Addition and subtraction of integral
Merging similar items is the key and difficult point. When merging similar items, we should pay attention to the following three points: 1. Master the concept of similar items, be able to distinguish similar items, and accurately master the two criteria for judging similar items: letter and letter index; 2. Make it clear that the meaning of merging similar items is to merge the similar items in a polynomial into one item. After merging similar items, the number of items in a polynomial will be reduced, (3) "combination" refers to the addition of the coefficients of the same category, and the results obtained are regarded as new coefficients, and the letters of the same category and the exponent of the letters should be kept unchanged
2. Multiplication and division of integers
The emphasis is on multiplication and division of integers, especially the multiplication formula. It is difficult for students to grasp the structural characteristics of multiplication formula and the extensive meaning of the letters in the formula. Therefore, the flexible use of multiplication formula is a difficulty. When adding brackets (or removing brackets), the handling of symbols in brackets is another difficulty. Adding brackets (or removing brackets) is a deformation of polynomials, In general, the key to division is to divide a polynomial by a single term
The main types of the four operations are as follows:
(1) Four operations of monomials
This kind of question mostly appears in the form of multiple choice question and application question
(2) The operation of monomials and polynomials
This kind of questions mostly appear in the form of solving questions with strong skills
Given Tana = 2, find (1) 5sina + 3cosa / 2sina cosa (2) 3sin ^ 2A + cos ^ a
(1) 5sina + 3cosa / 2sina cosa (numerator denominator divided by COSA) = (5tana + 3) / (2tana-1) = (10 + 3) / (4-1) = 13 / 3 (2) 3sin ^ 2A + cos ^ 2A = 3 (1-cos & # 178; a) + cos & # 178; a = 3-2cos & # 178; a = 2sina / cosa = 2sina = 2cosasin & # 178; a + cos & # 178; a = 1
Let's take an example of converting heat energy into electricity
This is just a topic in my physics exercise book
This problem, solar energy is not heat to electricity?
There are also thermal power plants, right
That's what I learned