Видео

{x+x ➖ }+{x+x ➖}={x^2+x^2}=x^4 (x ➖ 4x+4)

(x ➖ 5x+5).

(ـ1+1=2) لا يوجد في الرياضيات 😂

(3)+(3) =6 (y ➖ 3x+3).,(xy ➖ 3xy+2).

(2 ➖ 1)^3^4 (2 ➖ 1)3^2^2 (1 ➖ 1)^3^1^2 ()3^2 (x ➖ 3x+2).

{x^3+x^3 ➖ }=x^6+ {x^2+x^2 ➖}=x^4 {x^6+x^4}=x^10 x^4^6 x^2^2^3^3.x^1^13^2 x^3^2( x ➖ 3x+2).1/8 1/2^3 1/1^1^2 /1^2 (x ➖ 2x+1).

(x ➖ 2x+2).

(x ➖ 1x+1) (x ➖ 2x+2) (x ➖ 3x+3)(x ➖ 4x+4) (x ➖ 1x+1) (x ➖ 1x+1) (x ➖ 2x+2) (x ➖ 2x+2) (x ➖ 1x+1) (x ➖ 2x+2) .

{6x+6x ➖ }+{6x+6x ➖}={12x^2+12x^2}=24x^2 12^12x^4 3^4^3^4x^4 1^2^2^3^2^2x2^2 1^1^1^1^1^1x1^2 x^1^2 (x ➖ 2x+1).

{9x+9x ➖ }+{9x+9x ➖ }={18x^2+18x^2}36x^4 18^18x^4 9^9^9^9x^4 3^2^3^2^3^2^3^2x^2^2 1^1^1^1^1^1^3^1x1^2 3x^2 (x ➖ 3x+2). 3^2^3^2^3^2 1^1^1^1^3^2 3^2 (x ➖ 3x+2).

2^{x+x ➖ }+{3y+3y ➖ }= 2^{x^2+6y^2}=2^6xy^4 =64xy^4 8^8xy^4 2^3^2^3xy^2^2 1^1^1^1^1^1xy^1^2 xy^12 (xy ➖ 2xy+1).;2^{3x+3x ➖ }{y+y ➖ }=2^{6x^2^y^2}=2^6^xy^4 64xy^4 8^8xy^4 2^32^3xy^2^2 1^1^1^1^1^1xy^1^2 xy^1^2 (xy ➖ 2xy+1).100^100^24 10^10^10^104^6 2^5^2^52^5^2^52^2^3^2.1^1^1^1^1^1^1^11^13^2 3^2 (y ➖ 3x+2). 8^8 2^32^3 1^1^1^1^1^2 1^2 (y ➖ 2x+1).

Why do these channels all restate every step 4 times?(Ok, twice but still...)

{x+x ➖ }+{1+1 ➖ }/{x+x ➖ }= {x^2+2}/x^2=2x^2/x^2 =2x (x ➖ 2x+2).

{5x+5x ➖ }+5{x+2x ➖ 2}=10x^2 5^{x^2+2}=5^2x^2 =10x^2 {10x^2+10x^2}= 20x^4 5^4x^4 5^12^2x^2^2 1^11^1x1^2 x^1^2 (x ➖ 2x+1).

{6x+6x ➖ }/{2x+2 x➖ }+{3x+3x ➖ }=12x^2/{4x^2+6x^2}=12x^2/10x^4 =1.2x^2 1.1x^2 1x^2 (x ➖ 2x+1).

it's all base 5 right from the beginning: left side: 5^(1/2 + 1/4 +1/8 ) right side: 5^(3*x) so, 7/8 = 3x - there you are!

(x ➖ 7x+7).

(5^1 ➖ 1/2)^9(1^1 ➖ 1/2^1)^3^2 (/2^1)^3^1 (2)^3 (x ➖ 3x+2)

What notebook program is this, and what tablet do you use?

(3x+3x ➖}+{7+7 ➖ }+{x+x ➖} {2+2 ➖ }={6x^2+14}{x^2+4}={20x^2+4x^2}= 24x^4 4^6x^2 4^3^2x^4 2^2^3^2x^2^2 1^13^1x^1^2 3x^2 (x ➖ 3x+2). ♥

7^1^3^1 1^1^3^1 3^1 (x ➖ 3x+1).

Thanks for the video :) I went with logarithms (log7(x^3) = log7(7^3)), which gave only x = 7 as a result, would you say it's an error, and if so why? :) (I am pretty rusty, didn't do maths for like 20 years, but I'm curious, does it mean every time we could get a polynomial we should and cannot use other methods, unless we miss some of the solutions?)

Bruh, so much writing, I got answer in 5 secs, don't overcomplicate😂

❤(2x+2x ➖}/{32+32 ➖}/{ 2x+2x ➖ }={4x^2+64}/4x^2= 68x^2/4x^2=17x^1 (x ➖ 17x+1).

8+4=12

3

(3^2) ➖ (2^3) (3^1) ➖ (1^1^2).(3) (2];(x ➖ 3x+2). (9)^2 ➖ (8)^2= {81 ➖ 64}= 17 (x ➖ 17x+1) .

(x ➖ 3x+3).

(16)+(16)+3=35:(4^4)+(4^4)+3y (2^2^2^2)+(2^2^2^2)+3y (1^1^1^1)+(1^1^1^2)+3y (2)+3y (xy ➖ 3xy+2).

X-2 = -+5, x=3 or 7

x^6 x^3^2 (x ➖ 3x+2). 10^10^44 2^52^5^4^11 1^1^1^1^4^11^1 2^2^1^1 1^2(x ➖ 2x+1).

(2^2+1)+(2^2 ➖ 1) (1^2) +(1^2) (x ➖ 2x+1). (x ➖ 2x+1).

(0.5)+(0.5)=1(b ➖ 0.5a+0.5) .(1)+(1)=2 (c ➖ 1b+1). (2)+(1) =3 (c ➖ 2a+1).

x^3(x ➖ 3x+3). 6^6 3^2^3^2 1^1^3^2 (x ➖ 3x+2).

x^64 x^4^16 x^4^4^4^ x^2^2^2^2^2^2x^1^1^1^1^1^2 x^1^2 (x ➖ 2x+1) x^6 x^3^2 (x ➖ 3x+2).

{x+x ➖}{1+1 ➖}/{x+x ➖},={ x^2+2 }/x^2 =2x^2/x^2 ,2x (x ➖ 2x+2).

Good video. Can I message you on WhatsApp?

I just solved with in 10 sec😂 . preparing for jee

I also did not know that it was this easy!!! May I ask if it is possible to ptactice Lambert form on thia problem???

I did not know that it was easy. I am also wondering, is it acceptable to categorize the answer as omega squared. I mean it is just a cube root id unity right???

👌. Excelent!!!

(x ➖ 3x+3)2^2 +(x ➖ 11x+1)^2^2 (x ➖ 3x+1)^1^1+(x ➖ 1x+1)^1^2 (x ➖ 3) ()^2 (x ➖ 3x+2).10^70^6 10^7^10^6 2^5^7^2^5^3^2 1^1^1^1^1^3^2 3^2 x ➖ 3x+2).

6^6 3^2^3^2 1^1^3^2 (x ➖ 3x+2).

5^10 5^2^5 1^2^1 2^1 (x ➖ 2x+1)

I fcking hate when you use this lamda BS

{x+x ➖ }=x^2 {x+x ➖ }=x^2 {x^2+x^2}= x^4 x^2^2 (x ➖ 2x+2).

{x^3+x^3 ➖ }=x^6,+{x^2+x^2 ➖}=x^4 {x^6+x^4}= x10 x^2^5 (x ➖ 5x+2)

(2)+(3) (y ➖ 3x+2) (6)^2=36 (6^6 ) (3^2^3^2).(3^1^1^2 (3^2) (xy ➖ 3xy+2).

3.

7^7 ➖ 6^8 1^1 ➖ 6^2^3 3^2^1^1 3^2 (x ➖ 3x+2).