A Nice Algebra Math Problem Solve By Math Tutor Susharma
HTML-код
- Опубликовано: 29 сен 2024
- Welcome to problemsolvieg, the place to be for all things math!
Learn how to solve an exponential equation, step by step tutorial.
math Olympiad exam. Lambert w function. solve exponential equation.
a nice math Olympiad algebra problem
Today I will kiss you tips and tricks to solve the given Olympiad Math questions in a simple and easy way.
learn how to prepare for math Olympiad fast!
This is an educational channel. The main purpose of this channel is to help students. Solutions of various math related problems are uploaded in this channel. you can use this channel to learn Math level, from primary level to High School
math Olympiad,
A Nice Algebra Math Problem Solve By Math Tutor Susharma
Lamba double function,
Olympiad mathematics,
Mathematics Olympiad
Lamba function
Olympiad maths
linear algebra
math
Olympiad Math
math tutor Susharma
Susharma Ghosh
math Olympiad exam
Maths Olympiad
math Olympics
maths olympiad questions
Mathematical Olympiad
Olympiad exam
w function
algebra
math Olympiad problems
Maths Olympiad exam
Olympiad
Olympiad exam for class 2
2012 theory
Ioqm
how to solve equations
how to solve equation
hardest math question
exponential equations
differential equations class 12
differential equations
black pen red pen
American math Olympiad
algebra grade 7
Ioqm preparation
Lectures
math geometry
maths problems
problem solving
math problem solve
equation solving
math solve
1, 2, 3 go
learn Math
exponent math problem
Olympic math
Premath
Problemas de logaritmos
Bmo 2024
Rmo maths olympiad
exponential equations
functional equations Olympiad
go wrong
hardest Math question
how to solve equation
how to solve equations
#math #matholympiad #equationsolving #problemsolving
Confusion: if a +ab + b = 18, how can a + b =18? Does this imply ab=0? a or b = 0? Not one of the choices in the above video????
If (b+1)(a+1)=19•1 =>
(b+1)= 19. and (a+1) = 1
b= 19-1. a= 1 - 1
b=18✓. a= 0✓
nevertheless if
(a+1) = 19 and (b+1) = 1
a = 19 - 1 b = 1 - 1
a = 18✓ b = 0✓
in any case "a" or "b" is zero
Therefore:
(a+b) = 18 if a=0 or b=0
(0+18) = 18 ; (18 + 0) = 18
*(a+1)(b+1) =19=1×19*
*(a+1=19và b+1=1) hoặc (a+1=1 và b+1=19)*
*(a=18, b=0) hoặc (a=0, b=18)*
*(a+1)(b+1) =19=(-1)(-19)*
*=>a+b=18*
*(a+1=-19và b+1=-1) hoặc (a+1=-1 và b+1=-19)*
*(a=-20và b=-2) hoặc (a=-2 và b=-20)*
*=>a+b=-22*
b = 19 / (a + 1) - 1, because 19 is a prime number =>
1. a + 1 = 19 => b = 0 => a + b = 18
2. a + 1 = -19 => b = -2 => a + b = -20
in het tweede geval: a+ b = -1 -19 -2 = -22
Yeah, right. Stupid mistake😮
a=17/2, b=1
Are here many possibilities such as
a = 18, b = 0
Awesome video❤❤❤
Awesome❤❤❤
Why 19 + 1 on RHS simply disappear
19 + 1 = 1 + 19, so is the same
@@henkn2 okay now understood
A=3,b=3
??? a + ab + b = 15
9
#8 #
good
Why a+1 can not be equal to ±9.5, and b+1=±2? And how about a+1=3 and b+1=6⅓? Etc. What a fucking method of solving equations? Are you sure you study algebra in America and not in Zimbabawa?
Bro does not know what "a and b are integers" means💀💀💀
@@pelledanasten1615 I believe it will be easier to understand the solution if we find possible values of (a,b) pair. Check out (a,b) values (0,18), (18,0), (-2,-22) and (-22,-2). In all these case a+ab+b will equal 18. You can derive these 4 pairs from the intermediate steps shown by the author. These 4 cases of (a,b) will lead to the soluton a+b = 18 or -22.
Read the question before writing a stupid comment
there are obviously (uncountable) infinitely many solutions without the restriction otherwise as (a+1)(b+1) = 19, so:
a + b = {a - 1 + 19/(a+1) ; a real\{-1}}
Zimbabawa?
@@pelledanasten1615, it's even funnier when you notice the racism.