Question

1. Rezolvati:
solutia ecuatiei x²+6x+9=16
solutia ecuatiei (x+2)²-4=0
solutia ecuatiei 2x+3/3=9/2x-3
2. Aratati ca ecuatiile urmatoare sunt echivalente:
x(2x+1)+(x+1)(x+2)=2(x²+1)+1
si
2(x+1)²-(x-2)(x+2)=27
Va rog, ajutati-ma!

Answer 1

1.x²+6x+9=16=>x²+6x+9-16=0 =>x²+6x-7=0
a=1,b=6,c=-7
delta=b²-4ac=36+28=64
x1=6+8\2=14\2=7
x2=6-8\2=-2\2=-1
S={7}

Answer 2

[tex]x^2+6x+9=16\\ (x+3)^2=16\\ |x+3|=4\\ \rightarrow x+3=4\Rightarrow x_1=1\\ \rightarrow x+3=-4\Rightarrow x_2=-7\\ S:x\in \{1;-7\}\\ \\ (x+2)^2-4=0\\ x^2+4x+4-4=0\\ x(x+4)=0\\ \rightarrow x_1=0\\ \rightarrow x_2=-4\\ S:x\in \{0;-4\} [/tex]
[tex]\dfrac{2x+3}{3}=\dfrac{9}{2x-3}\\ (2x+3)(2x-3)=27\\ 4x^2-9=27\\ 4x^2=36\\ x^2=9\\ |x|=3\\ S:x\in \{\pm 3\}\\ \\ x(2x+1)+(x+1)(x+2)=2(x^2+1)+1\\ 2x^2+x+x^2+x+2x+2=2x^2+4x+2+1\\ 3x^2+4x+2=2x^2+4x+3\\ x^2=1\Rightarrow x\in \{\pm 1\}\\ \\ 2(x+1)^2-(x-2)(x+2)=27\\ 2x^2+4x+2-x^2+4=27\\ x^2+4x+4=25\\ (x+2)^2=25\\ |x+2|=5\\ \rightarrow x+2=5\Rightarrow x_1=3\\ \rightarrow x+2=-5\Rightarrow x_2=-7\\ S:x\in \{3;-7\}\\ \text{Nu inteleg la ce te referi prin:"ecuatiile urmatoare sunt echivalente"}[/tex]