2x2 egal4 x2 égal 8 plus 2 vaut 10

C,est fait exprès pour avoir des commentaire,en voilà le mien Tom soyer

C,est fait exprès pour avoir des commentaire,en voilà le mien paresseux

(4.X^1/2)^2=(4.X)^1/2 ; 16.X=2.X^1/2; 8.X=X^1/2 ; 64.X^2=X ; solution 1, X=0 ; solution 2 if X#0 ; 64.X =1; X= 1/64

Можно все намного короче и быстрее решить .

2x2x2+2=10

Страшно подумать, сколько времени и бумаги потребовалось бы автору для решения примера на Бином Ньютона.😂

Х=2 ; х=1/2.

x=9/16=0.5625

3

X=0 verifica la ecuación..

Решение через ОДНО место. Все проще и намного

YOUR METHOD IS TOO COMPLICATED OK . HOW CAN YOU JUST IMAGINE 27 & 7 TO SOLVE THE PROBLEMS ? . GO & KEARN FROM OTHER ON - LINE TEACHERS B4 COMING ON.LINE .

[√(x+(√2))]² = 2 x + (√2) = 2 x = 2 - (√2)

8ˣ = x⁶ x = ⁸log(x⁶) x = (⁶⁄₃) • ²log(x) *x = 2 • ²log(x)* *If (x = 2), so* 2 = 2 • ²log(2) 2 = 2 • 1 *2 = 2* ............ _(match)_ *If (x = 4), so* 4 = 2 • ²log(4) 4 = 2 • 2 *4 = 4* ............ _(match)_

X = 2

(x^2)(x)=8 x=2

Х=2. It's easy

Простейшая задача. Это не уровень олимпиада. Самая обычная задача русской школы.

I don’t know why people have a problem with… like x+ y =9 And they start with the solution by writing the problem again…

Гоните какую то пургу, не кому не интересно

Nooooooo where is thé other x+ 2 you missed

Why introduce logs so late

Очень долгое решение, можно намного короче решить

x^n --> n roots 2x³ = 16 --> 2x³ = 3 roots

4 değer verdinde çözmesi 3 saniye sürdü

х=3.😂

Y^3+20/(Y^3+2)=7 Y = Cbrt[2] Y =Cbrt[3] Y=-0.5Cbrt[3]-0.5Surd[243,6]i Y=-0.5Cbrt[2]±0.5Surd[108,6]i Y=-0.5Cbrt[3]+0.5Surd[243,6]i It’s in my head.

big fan brooo,❤❤❤❤

Crazy

V2 (commonly also -V2).

(3x) ^ (1/2) = x sqrt(3x) = x (sqrt(3x)) ^ 2 = x ^ 2 3x = x^2 x ^2=3x x=3 let's verify sqrt(3 * 3) = 3 sqrt(3 ^ 2) = 3 3=3 RHS =LHS hence ,, X=3

3rd equation (2^x)^3+2^x= “0????”

(2*2)/(х*х)=х/2 2*2*2=х*х*х х=2

why 0:53 not get x = х=8√2

2. Что тут считать

2

Erreur

4.

Thank you. But it’s possible a little differently . (1) X^4=-4=4*{cos[pi+2*pi*n]+i*sin[pi+2*pi*n] }=4*e^[i*pi*(1+2*n) , n=0,1,2,3 ; (2) Xn=sqrt(2)*e^(i*[pi/4]*[1+2*n] ]=sqrt(2)*{ cos( [pi/4]*[1+2*n ])+i*sin( [pi/4]*[1+2*n] ) ; (3.0) Xo=sqrt(2)*{cos(pi/4)+i*sin(pi/4) }=1+i ; (3.1) X1=sqrt(2)*{cos(3*pi/4)+i*sin(3*pi/4) }=-1+i ; (3.2) X2= sqrt(2)*{cos(5*pi/4)+i*sin(5*pi/4) }=-1-i ; (3.3) X3=sqrt(2)*{cos(7*pi/4)+i*sin(7*pi/4)}=1-i !!!!!!😊)!!!! With respect, Lidiy

Sure!

It may be a lengthy explanation, but it does clarify mathematical facts for a beginner like me. Like the square root of zero being zero. I often wondered about what zero to the power1\2 would be because zero to the power zero is an indeterminate. Am I right? Thank you, Phil Cool Maths.

The solution is more complicated as the problem.

The use of RI (a³ - b³) is very nice, useful and clear. There is another way: euclidean division between polynomials. (x√x)² = 8 x²·(√x)² = 8 x²·x = 8 x³ = 8 x = ³√8 x = 2 /// other roots (by the way of euclidean division): • x³ = 8 => x³ - 8 = 0 => x³ + 0·x² + 0·x - 8 = 0 • x = 2 => x - 2 = 0 (x³ + 0·x² + 0·x - 8)/(x - 2) = (x² + 2x + 4) x² + 2x + 4 = 0 Δ = 2² - 4·1·4 = 4 - 16 = -12 = -2²·3 √Δ = ±i√(2²·3) = ±2i√3 • root #1: x = (-2 + 2i√3)/(2·1) = -1 + i√3 • root #1: x = (-2 - 2i√3)/(2·1) = -1 - i√3 /// final results: ■ x = 2 ■ x = -1 + i√3 ■ x = -1 - i√3 🙂

It’s in my head.

(xSqrt[x])^2=x x = -1 x= 0 x = 1

After the first exponent to 2 both sides we can get the answer x = 16/sqrt(2), which is the same number as 8*sqrt(2), thus all other steps are just redundant...

It is boring. Why do you solve very slowly and long.

X= 2.

2