Răspuns:
a) x∈{1; 2}; b) x∈{-3; 1/3}; c) x∈{12/11; 2}; d) x∈{-1/2; 6}
Explicație pas cu pas:
ca tripletul de numere a; b; c să fie termeni consecutivi ai unei progresii geometrice trebuie să se respecte proprietatea că b²=a·c
Atunci vom obţine:
a) (5x-8)²=(4x-7)(7x-10)⇔25x²-80x+64=28x²-40x-49x+70⇔3x²-9x+6=0 |÷3
x²-3x+2=0, ecuatie de gr.II cu soluţiile x=1 şi x=2.
b) (2+x)²=4·(x²+3x+(1/4))
4+4x+x²=4x²+12x+1⇔3x²+8x-3=0, ecuatie de gr.II
Δ=8²-4·3·(-3)=64+36=100>0
[tex]x_{1}=\frac{-8-\sqrt{100} }{2*3} =\frac{-8-10}{6}=\frac{-18}{6} =-3\\ x_{2}=\frac{-8+\sqrt{100} }{2*3} =\frac{-8+10}{6}=\frac{2}{6} =\frac{1}{3} \\[/tex]
x∈{-3; 1/3}
c)
[tex](3x-4)^{2}=\frac{x}{3} *(5x-4)\\[/tex]
3·(9x²-2·3x·4+4²)=x·(5x-4)⇔27x²-72x+48=5x²-4x⇔22x²-68x+48=0|÷2
11x²-34x+24=0
Δ=(-34)²-4·11·24=1156-1056=100>0
[tex]x_{1}=\frac{34-\sqrt{100}}{2*11} =\frac{24}{22}=\frac{12}{11}\\ x_{2}=\frac{34+\sqrt{100} }{2*11} =\frac{44}{22}=2.\\[/tex]
Deci x∈{12/11; 2}
d) (√(9x+6))²=(x-1)·2x⇔9x+6=2x²-2x⇔2x²-2x-9x-6=0⇔2x²-11x-6=0
Δ=(-11)²-4·2·(-6)=121+48=169>0
[tex]x_{1}=\frac{11-\sqrt{169} }{2*2} =\frac{11-13}{4}=\frac{-2}{4} =-\frac{1}{2}\\ x_{2}=\frac{11+\sqrt{169} }{2*2} =\frac{11+13}{4}=\frac{24}{4} =6\\[/tex]
Deci x∈{-1/2; 6}
