Ask a high school math solid geometry problem. There are eight items, namely, city AB and AD midpoint connection. BC, DC midpoint connection. A 1B 1 midpoint connection. D 1c 1 midpoint connection. AB, A 1B 1 midpoint connection.
A high school math problem, to be solved! Hope to adopt!
Analysis: according to the difference between two digits is 56, list X-Y = 56; According to the fact that the last two digits of the square of two digits are the same, x2-y2=m× 100(m is a positive integer) is obtained; Solve the equations, deduce the value of m, and then get the value of y.
Answer: ∫x-y = 56, x2-y2=m× 100(m is a positive integer),
X elimination gives112y =100m-3136, y=(25m/8)-28,
∫y is a two-digit number, m < 100,
∴m=56 or 84,
Y = 22 or 47.
When y=22, x = 78.
When y=47, x= 103 (omitted).
So the answer is: 22,78.
Find the solution of a high school mathematical equation X/E X = E T+T.
e^x=x/(e^t+t)
x=ln[x/(e^t+t)]=lnx-ln(e^t+t)
X is a transcendental equation because it appears in both exponential and linear terms.
What can't be solved under normal circumstances,
If it only appears in the index, it can be expressed by logarithm.
Find the solution of a high school math problem, and the main flow is set to 2x = t(t >；; 0)
The original equation is t? +at+a+ 1=0
If there is a real root, △=a? -4(a+ 1)≥0
A≤2-2√2 or a≥2+2√2
Let f(t)=t? +at+a+ 1
The symmetry axis is t=-a/2.
when-a/2 & gt； 0 is a
When -a/2≤0, that is, a≥0, △≥0 and f (0) are satisfied.
The range of a is (-∞, 2-2√2).
How to solve a high school math problem without doing it in detail?
A high school math problem (geometric proof) passes through E, makes AD parallel lines, and passes through DC to G, then EG: AD = 1: 3.
CG:DG= 1:2，
So DG=2/3DC=2/3BD,
So FD: eg = 3: 5,
FD = 3/5EG =(3/5)*( 1/3)AD = 1/5AD，
So af: FD = 4: 1.
When solving mathematical geometry problems in senior high school, good people can give a special example, that is, a regular hexagon, which is equivalent to the ratio of the area of a regular hexagon with a side length of A to the area of a regular hexagon with a side length of 2a, so it is 1: 4.
A high school mathematics derivative problem, the solution of x∈[2, ∞), f(x)≥0, that is, x? +3ax？ +3x+ 1 >=0, that is, x+3/x+ 1/x? & gt=-3a
X∈[2, ∞), -3a.
Let g(x)=x+3/x+ 1/x?
g'(x)= 1-3/x？ -2/x？ =(x？ -3x-2)/x？
Let's prove that G' (x) >: =0 holds in x∈[2, ∞), that is, x? -3x-2 >=0 holds in x∈[2, ∞).
Let h(x)=x? -3x-2；
h'(x)=3x？ -3=3(x+ 1)(x- 1), and it is easy to know h' (x) >1; 0 is always true in x∈[2, ∞], so g(x) is increasing function in x∈[2, ∞], so H (x) >1; =h(2)=0, which means x? -3x-2 >=0 holds in x∈[2, ∞),
That is g' (x) >; =0 is a constant in x∈[2, ∞], and g(x) is a increasing function in x∈[2, ∞].
So the minimum value of g(x) is g(2)= 15/4,
So-3a
Get an a & gt=-5/4.
Solve a high school math problem. Convert an equation into a standard equation of a circle
That is, (x+m/2) 2+(y+n/2) 2 = (m/2) 2+(n/2) 2.
If the center coordinate is (2,-1), then -m/2=2 -n/2=- 1.
So m=-4 n=2.
r^2=(m/2)^2+(n/2)^2=5
Ji r= radical number 5