Salut!
Voi da doar raspunsurile, explicatiile le voi scrie unde este nevoie.
a)
[tex]\left(5-3\right)^2=5^2+\left(-3\right)^2+2x\cdot \:5x\left(-3\right)[/tex]
[tex]2x\cdot \:5x\left(-3\right)=-30[/tex]
(am schimbat membrul stang cu membrul drept si am scazut 5^2+(-3)^2 din ambele parti)
[tex]-30x^2=-30[/tex]
[tex]x^2=1[/tex]
x = ±1
b)
[tex]\left(-7\right)^2+\left(2\right)^2+2x\left(-7\right)x\left(2\right)=\left(-7+2\right)^2[/tex] (acelasi concept ca mai sus)
[tex]2x\left(-7\right)x\cdot \:2=-28[/tex]
[tex]-28x^2=-28[/tex]
[tex]x^2=1[/tex]
x = ±1
c)
[tex]\left[-7-\left(-3\right)\right]x\left[-7+\left(-3\right)\right]=\left(-7\right)^2-\left(-3\right)^2[/tex]
[tex]\left[-7-\left(-3\right)\right]x\left[-7+\left(-3\right)\right]=40[/tex]
[tex]\frac{40x}{\left(3-7\right)\left(-7-3\right)}[/tex] [tex]=\frac{40}{\left(-7+3\right)\left(-7-3\right)}[/tex]
[tex]\frac{40x}{40} = \frac{40}{40}[/tex]
x = 1
d)
[tex]\left(-11-10\right)x\left(-11+10\right)=\left(-11\right)^2-10^2\\[/tex]
[tex]\left(-11-10\right)x\left(-11+10\right)=21[/tex]
[tex]\frac{21x}{21}[/tex] [tex]=\frac{21}{21}[/tex]
x = 1
