Ec. Are 6 soluții prin permutarea soluției (2 ,3,4 ) PTR.. in enunț nu se specifica nimic despre x, y,z

If we squaring both sides several times then we get the value of x. This alternative method also a good idea.Thank you sir!

All maths lovers are requested to watch this mathematics.This mathematics is very important for Olympiad Question also competitive exams. We should learn this mathematics.

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I would start withh rewritung 7^(x+5) to 7^(2+(x+3)): 7^(x+3)+7^(x+5)=50 7^(x+3)+7^(2+(x+3))=50 7^(x+3)+7^2*7^(x+3)=50 7^(x+3)*(1+7^2)=50 7^(x+3)*(49+1)=50 7^/x+3)*50=50 7^(x+3)=1 7^(x+3)=7^0 x+3=0 x=--3

(9^2001-9^1999)/(3^4000-3^3996)= (9^2001-9^1999)/((3^(2*2000)-3^(2*1998))= | 4000=2*2000 and 3996=2*1998 (9^2001--9^1999)/((3^2)^2000-(3^2)^1998)= | a^(m*n)=(a^m)^n (9^2001--9^1999)/(9^2000-9^1998)= | 3^2=9 (9^(2+1999)-9^1999)/(9^(2+1998)-9^1998)= | a^(m+n)=a^m*a^n (9^2*9^1999--1*9^1999)/(9^2*9^1998-1*9^1998) | 9^1999 is common in enumerator and 9^1998 is common in denominator 9^1999*(9^2-1)/(9^1998*(9^2-1)= | (9^2--1) is common in enumerator and denominator 9^1999/9^1998= | a^m/a^n=a^(m-n) 9^(1999-1998)= 9^1= 9 I think, it is easier to take 9^1998 out of the sum in the enumerator and to take out 9^1996 out of the sum in the denominator If you do that, you would not get fractions divided b fractions like in the video.

Nice maths! Nice 👍 solution

x=2;y=3;z=4 x=2;y=4;z=3 x=3;y=2;z=4 x=3;y=4;z=2 x=4;y=2;z=3 x=4;y=3;z=2

Very interesting method of solving

(x-4)^4=3^4 (x-4)=3. X=3+4. x=7

Wrong aproach

And degrees

🌹Chersy Khan k kamalaat dekhain…. Siri Lanka tu na bana sakka… magar Mulk aur qoom ( Choutiaz) ka bairah gharak kar diya…🧐

Excellent

All are requested to learn this mathematics!

336 = 16.21 = 16.(16 + 4 + 1) = (4^2).(4^2 + 4^1 + 4^0) = 4^4 + 4^3 + 4^2, then a = 4, b = 3, c = 2 (or other order)

Where n is a member of Z ... very important to mention too ... thank you sir ...

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Clever! Dr. Ajit Thakur (USA).

Another great demonstration Chacha ! Thanks for sharing Man !

u can rewrite 8^(x-8) into (1/4)^(-3/2(x-8)), you can then simplify into 4-x = -3x/2 +12, then with some basic algebra you get x=16.

(First: x <>0 mod Pi/2 We also notice that tan(Pi/8) = sqrt(2) - 1 and that cotan(Pi/8) = sqrt(2) +1) We have (1/tan(x)) -tan(x) = 2 or (tan(x)^2 +2.tan(x) -1 = 0 Then tan(x) = sqrt(2) - 1 or tan(x) = -(sqrt(2) +1) *If tan(x) = sqrt(2) - 1 = tan(Pi/8) then x = Pi/8 mod Pi *If tan(x) = -(sqrt(2) - 1) = -cotan(Pi/8) = -tan(Pi/2 - Pi/8) = tan(-3.Pi/8) then x = -3.Pi/8 mod Pi *We can sum up: x = Pi/8 mod Pi/2 (as -3.Pi/8 = Pi/8 - Pi/2)

Merci monsieur

সহজ করে পড়িয়েছেন। অঙ্কটা বুঝতে পারলাম। ধন্যবাদ স্যার।

Nice solution 👌 👍

3998(3*4-1)/3996(3*4-1)cancel3*4-1 3*3998-3996 equals 3*2 is 9

Simplify to (3^4002 - 3^3998)/(3^4000-3^3996), you can then factor out a 3^2 on top, getting 3^2(3^4000-3^3996)/(3^4000-3^3996), which is = 3^2 = 9

Nice trick

Ok, understood the mathematics

Great. Thank you, sir.

This mathematics is for class x students of all countries.They are requested to learn this mathematics.

Easier factoring up and down: 3^4002 ( 1 - 3^(-4) ) / 3^4000 (1 - 3^(-4)) = 3^4002 - 3^4000 = 3^2 = 9 , simpler and elegant.

Verified, The roots satisfied with the original equation. Thank you for solution!

30 сек на решение устно!

এই অঙ্কটা আগে থেকেই জানি। আরেকবার দেখার সুযোগ পেলাম। খুব ভালো।

Very helpful sir. Thank u 👍

He has unnecessarily complicated the solution. The problem could have been solved in a much simpler way.

Glad to see.

Nice sir

X equals to 1/5 not satisfying the equation. Please explain.

Excellent,again well done Thank you.

Let n=x^½ 7ⁿ+7ⁿ+7ⁿ=63ⁿ 3(7ⁿ)=63ⁿ 9ⁿ=3 3²ⁿ=3¹ 2n=1 n=½ x^½=½ x=¼ ❤

Ok nice 👍 solution

It will be sinx instead of six.please do the needful.

Reemplazo equivalencias de variables en ecuación y obtengo z= 3 y=. 8. x= 5

3 veces 7^√× = 63^√× 3=. (63/7)^√×. X. = 2

An Important Olympiad Mathematics: 4^x + 4^y + 4^z = 336; x < y < z = ? 4^x + 4^y + 4^z = 336 = (16)(21) = 4²(1 + 4 + 16) = 4² + 4³ + 4⁴ x = 2, y = 3, z = 4 Answer check: 4^x + 4^y + 4^z = 336; Confirmed as shown Final answer: x = 2, y = 3, z = 4

MATH FOR ALL: 1/x = 3125ˣ; x = ? x ≠ 0; (1/x)¹⸍ˣ = (3125ˣ)¹⸍ˣ = 3125 = 5⁵, 1/x = 5; x = 1/5 Answer check: 1/x = 1/(1/5) = 5, 3125ˣ = (5⁵)¹⸍⁵ = 5; Confirmed Final answer: x = 1/5

A Nice Radical International Olympiad Mathematics: 7^√x + 7^√x + 7^√x = 63^√x; x = ? 3(7^√x) = [(9)(7)]^√x = [(3²)^√x](7^√x) = (3^2√x)(7^√x), 7^√x > 0 3^2√x = 3 = 3^1, 2√x = 1, √x = 1/2; x = 1/4 Answer check: 7^√x + 7^√x + 7^√x = 3[7^(1/2)] = 3√7, 63^√x = √63 = 3√7; Confirmed Final answer: x = 1/4

x^2 = (4)(6)(8)(10) + 16 = (16)(120 + 1) = (4^2)(11^2) = 44^2; x = +/- 44