Ecuația de forma x² = a, unde a ∈ R
E.682. Rezolvați ecuațiile:a) x 2 = 400 ; x^2 = 400;\qquad x 2 = 400 ; b) 2 x 2 = 0 ; 2x^2=0;\qquad 2 x 2 = 0 ; c) x 2 + 9 = 0 x^2+9=0 x 2 + 9 = 0
Răspuns: a) x = ± 20 ; x=\pm20; x = ± 20 ; b) x = 0 ; x=0; x = 0 ; c) x = ϕ ; x=\phi; x = ϕ ;
Soluție:
Observăm că o ecuație de forma x 2 = a , x^2=a, x 2 = a , a ∈ R a\in \R a ∈ R
E.683. Rezolvați ecuațiile:a) x 2 = 256 ; x^2 = \sqrt{256};\qquad x 2 = 256 ; b) ; 29 ( x 2 − 3 ) = 0 ; ;\qquad 29(x^2-3)=0;\qquad ; 29 ( x 2 − 3 ) = 0 ; c) 2 x 2 = 1 , 62. 2x^2=1,62. 2 x 2 = 1 , 62.
Răspuns: a) x = ± 4 ; x=\pm4; x = ± 4 ; b) x = ± 3 ; x=\pm3; x = ± 3 ; c) x = ± 9 10 ; x=\pm\dfrac{9}{10}; x = ± 10 9 ;
E.684. Rezolvați ecuațiile:a) 7 ( x 2 + 2 ) = 98 ; 7(x^2+\sqrt{2})=\sqrt{98};\qquad 7 ( x 2 + 2 ) = 98 ; b) ( x + 3 ) 2 = 27 ; (x+\sqrt{3})^2=27;\qquad ( x + 3 ) 2 = 27 ; c) 4 x 2 = ( 2 + 3 ) 2 4x^2=(2+\sqrt{3})^2 4 x 2 = ( 2 + 3 ) 2
Soluție:
a) 7 ( x 2 + 2 ) = 7 2 7(x^2+\sqrt{2})=7\sqrt{2} 7 ( x 2 + 2 ) = 7 2 x 2 + 2 = 2 ⇒ x = 0 . x^2+\sqrt{2} = \sqrt{2} \Rightarrow \boxed{x=0}. x 2 + 2 = 2 ⇒ x = 0 .
b) x + 3 = ± 3 ⇒ x 1 = 0 ; x 2 = − 2 3 . x+\sqrt{3}=\pm\sqrt{3} \Rightarrow \boxed{x_1=0;~ x_2=-2\sqrt{3}}. x + 3 = ± 3 ⇒ x 1 = 0 ; x 2 = − 2 3 .
c x 2 = ( 2 + 3 2 ) 2 ⇒ x = ± 2 + 3 2 . x^2=\Big(\dfrac{2+\sqrt{3}}{2}\Big)^2 \Rightarrow \boxed{x=\pm\dfrac{2+\sqrt{3}}{2}}. x 2 = ( 2 2 + 3 ) 2 ⇒ x = ± 2 2 + 3 .
E.685. Rezolvați ecuațiile:a) 1 + ( x + 1 ) 2 = 0 ; 1+(x+1)^2=0;\qquad 1 + ( x + 1 ) 2 = 0 ; b) 9 − x 2 16 = − 1 ; \dfrac{9-x^2}{16}=-1;\qquad 16 9 − x 2 = − 1 ; c) 4 4 − x 2 = 1 2 ; . \dfrac{4}{4-x^2}=\dfrac{1}{2};. 4 − x 2 4 = 2 1 ; .
Răspuns: a) x = ϕ ; x=\phi; x = ϕ ; b) x = ± 5 ; x=\pm5; x = ± 5 ; c) x = ϕ ; x=\phi; x = ϕ ;
E.686. Arătați că următoarele perechi de ecuații au aceleași soluții:a) ∣ x + 2 ∣ = 5 |x+2|=5 ∣ x + 2∣ = 5 x 2 + 4 x + 4 = 25 ; x^2+4x+4=25;\qquad x 2 + 4 x + 4 = 25 ; b) ∣ x − 2 ∣ = 3 |x-2|=3 ∣ x − 2∣ = 3 ( x − 2 ) 2 = 9. \sqrt{(x-2)^2}=9. ( x − 2 ) 2 = 9.
E.687. Rezolvați în N \N N a) x 2 + 2 y 2 = 9 ; x^2+2y^2=9;\qquad x 2 + 2 y 2 = 9 ; b x + 1 x + 2008 y 2 = 2010. x+\dfrac{1}{x}+2008y^2=2010. x + x 1 + 2008 y 2 = 2010.
E.688. Rezolvați în N \N N a) ( x + 1 ) + ( x + 2 ) + … + ( x + 50 ) = ( x + 25 ) 2 + 25 ; (x+1)+(x+2)+ \ldots + (x+50)=(x+25)^2+25; ( x + 1 ) + ( x + 2 ) + … + ( x + 50 ) = ( x + 25 ) 2 + 25 ; b) 1 + 3 + … + ( 2 x − 1 ) = 100. 1+3+ \ldots + (2x-1)=100. 1 + 3 + … + ( 2 x − 1 ) = 100.
E.689. Rezolvați ecuațiile:a) 2 x 2 = 3 3 − 3 2 : ( 2 6 − 2 5 − 2 4 − 2 3 − 2 2 − 2 1 − 2 0 ) ; 2x^2=3^3-3^2:(2^6-2^5-2^4-2^3-2^2-2^1-2^0); 2 x 2 = 3 3 − 3 2 : ( 2 6 − 2 5 − 2 4 − 2 3 − 2 2 − 2 1 − 2 0 ) ; b) x 2 = 2 ⋅ 2 2 ⋅ 2 2 2 : [ 2 + 2 2 + 2 2 2 : ( 2 2 2 − 2 2 ⋅ 2 ) ] . x^2=2\cdot 2^2\cdot 2^{2^2}:\big[2+2^2+2^{2^2} :(2^{2^2}-2^2 \cdot 2)\big]. x 2 = 2 ⋅ 2 2 ⋅ 2 2 2 : [ 2 + 2 2 + 2 2 2 : ( 2 2 2 − 2 2 ⋅ 2 ) ] .
E.690. Rezolvați ecuațiile:a) x 2 = 2009 ⋅ 2011 + 1 ; x^2=2009 \cdot 2011 + 1; x 2 = 2009 ⋅ 2011 + 1 ; b) x 2 = 2009 ⋅ 2010 ⋅ 2011 ⋅ 2012 + 1. x^2=2009 \cdot 2010 \cdot 2011 \cdot 2012 + 1. x 2 = 2009 ⋅ 2010 ⋅ 2011 ⋅ 2012 + 1.
Soluție:
a) x 2 = ( 2010 − 1 ) ( 2010 + 1 ) + 1 = 2010 2 ⇒ x = ± 2010 . x^2=(2010-1)(2010+1)+1 = 2010^2 \Rightarrow \boxed{x= \pm2010}. x 2 = ( 2010 − 1 ) ( 2010 + 1 ) + 1 = 201 0 2 ⇒ x = ± 2010 .
b) Obs : Este mai ușor de rezolvat pe cazul general:x 2 = n ( n + 1 ) ( n + 2 ) ( n + 3 ) + 1 , n ∈ Z . x^2=n(n+1)(n+2)(n+3)+1,~ n \in \Z. x 2 = n ( n + 1 ) ( n + 2 ) ( n + 3 ) + 1 , n ∈ Z . x 2 = [ n ( n + 3 ) ] ⋅ [ ( n + 1 ) ( n + 2 ) ] = x^2 = [n(n+3)] \cdot [(n+1)(n+2)]= x 2 = [ n ( n + 3 )] ⋅ [( n + 1 ) ( n + 2 )] = = ( n 2 + 3 n ) ( n 2 + 3 n + 2 ) + 1 = =(n^2 + 3n)(n^2 + 3n + 2)+1= = ( n 2 + 3 n ) ( n 2 + 3 n + 2 ) + 1 = = p c t . a ( n 2 + 3 n ) 2 ⇒ x = ± ( n 2 + 3 n ) . \overset{pct.a}{=}(n^2+3n)^2 \Rightarrow \boxed{x=\pm(n^2+3n)}. = p c t . a ( n 2 + 3 n ) 2 ⇒ x = ± ( n 2 + 3 n ) .
E.691. Rezolvați ecuația x + y + z + 8 = 2 ( x − 1 + 2 y − 2 + 3 z − 3 ) , x+y+z+8=2(\sqrt{x-1}+2\sqrt{y-2}+3\sqrt{z-3}), x + y + z + 8 = 2 ( x − 1 + 2 y − 2 + 3 z − 3 ) , x , y , z ∈ R , x,y,z \in \R, x , y , z ∈ R , x ≥ 1 , y ≥ 2 , z ≥ 3. x\geq 1, ~y\geq 2, ~z\geq 3. x ≥ 1 , y ≥ 2 , z ≥ 3.
Soluție:
Ecuația se mai scrie:
[ ( x − 1 ) − 2 x − 1 + 1 ] + [ ( y − 2 ) − 4 y − 2 + 4 ] + [ ( z − 3 ) − 6 z − 3 + 9 ] = 0 [(x-1) - 2\sqrt{x-1} + 1] + [(y-2) - 4\sqrt{y-2} + 4] + [(z-3) - 6\sqrt{z-3} + 9] = 0 [( x − 1 ) − 2 x − 1 + 1 ] + [( y − 2 ) − 4 y − 2 + 4 ] + [( z − 3 ) − 6 z − 3 + 9 ] = 0 ( x − 1 − 1 ) 2 + ( y − 2 − 2 ) 2 + ( z − 3 − 3 ) 2 = 0. (\sqrt{x-1}-1)^2 + (\sqrt{y-2}-2)^2 + (\sqrt{z-3}-3)^2=0. ( x − 1 − 1 ) 2 + ( y − 2 − 2 ) 2 + ( z − 3 − 3 ) 2 = 0. 0 , 0, 0 , x = 2 , y = 6 , z = 12 . \boxed{x=2,~y=6,~z=12}. x = 2 , y = 6 , z = 12 .