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1INTERESTING EQUATIONS AND INEQUALITIES
1Ibragimov A., 2Pulatov O., 3Qochqarov A.
1Professor 2Assistent 3Student
1,2,3Uzbekistan-Finland Pedagogical Institute https://doi.org/10.5281/zenodo. 10196007
Abstract. In this article, a new method is used to solve some equations and inequalities, and it is important for students to learn differential calculus. The use of differential calculus allows to solve complex equations and inequalities in a simple way, and this method can also be used to solve some geometric problems.
Keywords: equation, inequality, triangle, rectangle, center, radius, bisector.
The main focus of this topic is that a new method is used to solve equations and inequalities, and in addition to students learning differential calculus, the use of differential calculus not only allows solving complex equations and inequalities in a simple way, but also some geometrical methods from this method can also be used to solve problems.
1) Let's assume . ..,Qn (1). If (1) are integers corresponding to Eq.
(x — a1)(x — a2) ... (x — am) — (—1)n • (n!)2 = 0 We find r that has an entire solution
Ql+<?2+"+<?2n 2n
Solving: r^ aj it is seen.
[(r — ai)(r — a2) ... (r — a2„)] > |(1)(2) ... (n) • (—1)(—2) ... (—n)| = (n!)2 For Eq
(r — a1,r — a2 ,r — a2n} = {1,2, ..., n, —1, —2,..., (—n)} should be therefore
(r — a1)(r — a2) ... (r — a2n) = 2nr — (a1 + a2 + —+ a2n) =
= 2nr — a1+.. +a2n 1+2+.. +n + (—1) + (—2)+.. +(—n) = 0 = r = .ai+a2+-+a2"
2n
2) If tx t2 _ tn for the numbers
n2 + 1 > (tx,t2.....tn)(—+ — +,..,+ —)
V tx t2 tn/
i 2 <- n'
If the relationship is appropriate, then 1<i<y<fc<n [7] if the inequality is true V /,_/, k for s tj,ty,tfc prove that the numbers are the lengths of the sides of a triangle.
Proof: Let's assume that the numbers t1,t2,t3 satisfying the condition of the problem cannot be the lengths of the sides of a triangle, i.e. t3 + t2 < t1 so be it
( (2.....tn) + = n + + + + + " + + W"-" >
\ti t2 t^/ ^^2 ^ t3 ti/ \tn—i tn /
n +---+ t
n+
n
1 H t2 (3/
if^(i271 + ti) + n(n-i)
u-u\u-u 3 2/
+ 2 + 2 + -2=>
4 = n2
4+
• ts V ti V
( 2 + 3)2
2 -4+ 1 + —-— >
2 • 3
> n2 - 3 +
4 2 • 3
= l + n; because t3 + t2 < t1 was.
We are conflicted. That is, our hypothesis is wrong. The issue has been completely resolved.
3) ABCD- Let the circle be an inner rectangle. D from the point BC, BA and AB The bases of the perpendicular drawn to the sides are respectively R,Q,R if. PQ=QR to be ABC and ADC bisectors of angles AC
Prove that it is necessary and sufficient that it intersects in a straight line. Proof: According to Simpson's theorem Q,P,R points lie on a straight line. CQD+CDP=180° and ARD=AQD=90°=>PCQD RADQ circle radii respectively ^ =
and = ^ is equal to. [4]
m
In that case PQ=2«1 sin QCP = 2-^smfr - 4SC) = CD • sin^CD
4) a,b,c belongs to ß+ and abc=1 If so, prove it.
5 =
l
+
l
+
l
3 > -
a3(b + c) b3(c + a) c3(a + b) 2 Proof: S- ( a(b + c) + b(c + a) + c(a + b)) > (1 +1 +1)2 = (ab + be + ac)2
((aö+öc+ac))2
=> 5 >
aö+öc+ac 33Va2ö2c2 _ 3
2 > 2 =2
5) If a,b,c belongs to R for ab + be + ac = 0 if P(a-b)+P(b-c)+P(c-a)=2P ( a + b + c)
Find all polynomials with real coefficients R(x) satisfying the condition. Proof: a=b=0=> P(0)+P(-c)+P(c)=2P(c)=P(c)= P(0)+P(-c)=P(0)=0=P(c)=P(-c)=P(x) couple So P(x) is a polynomial.
P(x)=anx2n + anx2n 2 + —+ a2x2n x+..+axx2 (a belongs to R, an Ф 0) in appearance
a=x b=2x c=-2 (x€Ä) if ab + be + ac = 2x2 — —— — = 0 So
3 v y 3 3
u 8 5 , ъ , 7
b-c=-x, c-a=—x, a+b+c=-x
3 3 3
P(-x)+P(|x) + P (— 5x) = 2P (7x) = (x2n ifwe equate the coefficients in front of)
an ((-1)2n + (I)2" + g)2") = 2an (7)2n = 32" + 82n + 52" =2^ 72" = 82n < 2 • 72n = (8)2" < 2; (2)
But n> 3 da (8)2n (i+1)2n > 1 + C2in1 + C|n- = 1+2n + ni2n-i2>
\7J \ 7J 2n 7 2n 49 7 49
— >2 So (*) the inequality is valid for n< 2 in the form of P(x)= a2x4 + a1x2 as a result of
42
direct inspection P(x)= a2x4 + a1x2, a2 a1€R it is not difficult to see the multiple issue condition. So the result P(x)= a2x4 + a1x2 (Va2,a1€R).
6) Let's say AB^AC in acute angle ABC. A circle drawn as a diameter with side BC intersects sides AB and AC at points M and N, respectively. If we mark the middle of the side BC with O, then the bisectors of the angles BAC and MON intersect at the point R, prove that the circles drawn outside the angles BM and CNR intersect on the side BC.
Proof: AMOR=ANOR because MO=NO,MOR=NOR, RO common side. So MR=NR;
BMCN cyclicity of the rectangle BCA=AMN ABC=ANM: according to the condition AB* 4C, 45C * 4CS => 4MN * 4NM (2)
AMRA and ANRA the radius of the outer circle.
Conclusion
In conclusion, from what has been analyzed and discussed above in the proving process it can be inferred that chosen formula has been proved with its clear process of checking and calculation which has been intended to do.
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