Question

urgent la matematică dau totul . Clasa a X a​

Answer 1

$\it 1.\ \ \sqrt{2x+3}=3 \Rightarrow (\sqrt{2x+3})^2=3^2 \Rightarrow 2x+3=9|_{-3}\Rightarrow 2x=6|_{:2} \Rightarrow x=3\\ \\ 2.\ \ \sqrt{x^2+1}=x+1 \Rightarrow (\sqrt{x^2+1})^2=(x+1)^2 \Rightarrow \\ \\ \Rightarrow x^2+1=x^2+2x+1|_{-(x^2+1)} \Rightarrow 0=2x \Rightarrow x=0\\ \\ 4.\ \ \sqrt{x^2+2x+1}=2 \Rightarrow \sqrt{(x+1)^2}=2 \Rightarrow |x+1|=2 \Rightarrow x+1=\pm1 \Rightarrow \\ \\ \Rightarrow x+1\in\{-1,\ 1\}|_{-1} \Rightarrow x\in\{-2,\ 0 \}\Rightarrow x_1=-2,\ \ x_2=0$

$\it 3.\ \ \sqrt{x-1}=x-3 \Rightarrow (\sqrt{x-1})^2=(x-3)^2 \Rightarrow x-1=x^2-6x+9 \Rightarrow \\ \\ x^2-7x+10=0 \Rightarrow x^2-2x-5x+10=0 \Rightarrow x(x-2)-5(x-2)=0 \Rightarrow \\ \\ \Rightarrow (x-2)(x-5)=0 \Rightarrow x\in\{2,\ 5\}\\ \\ Verificare\ (obligatorie) :\\ \\ x=2 \Rightarrow \sqrt{2-1}=2-3 \Rightarrow 1=-1\ (F)\ \Rightarrow x=2\ nu\ e\ solu\c{\it t}ie\ pentru\ ecua\c{\it t}ia\ dat\breve a\\ \\ x=5 \Rightarrow \sqrt{5-1}=5-3 \Rightarrow 2=2\ (A)$

Deci, ecuația dată admite o soluție unică, x=5.