Question

Cum se rezolva ecuatiile cu radicali, de genul:
x√2=4
sau
y√2+4=y+ 4√2
sau
2√3 - 3=x√3

Answer 1

$x \sqrt{2}=4=>x= \frac{4}{ \sqrt{2} }= \frac{ 4\sqrt{2} }{2}= 2\sqrt{2}. \\ \\ y \sqrt{2}+4=y+4 \sqrt{2}<=> y\sqrt{2}- y=4\sqrt{2}-4<=> \\ <=>y( \sqrt{2}-1)=4( \sqrt{2}-1)=>y=4. \\ \\ 2\sqrt{3}-3= x\sqrt{3}=>x= \frac{ 2\sqrt{3}-3 }{ \sqrt{3} }= \frac{ \sqrt{3}(2 \sqrt{3}-3) }{3}= \frac{6- 3\sqrt{3} }{3}=2- \sqrt{3}.$

Answer 2

x√2=4
x=4÷√2
x=4√2/2
x=2√2

y√2+4=y+4√2
y√2-y=4√2-4
y(√2-1)=4(√2-1)
y=4(√2-1)/√2-1
y=4(√2-1)(√2+1)/2-1
y=4(√2^2-1^2)
y=4x2-4
y=4

2√3-3=x√3
x=2√3-3/√3
x=(2√3-3)ori√3/3
x=2√3√3-3√3/3
x=2.ori 3-3√3/3
x= 3(2-√3)/3
x=2minus √3