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Determinați numerele naturale a,b,c,d dacă :

a+1/b+2=1/2 ; b+2/c+3=2/3 ; c+3/d+4=3/4 și a²+b²+c²+d²=270.​

Răspuns:

[tex]\frac{a+1}{b+2}=\frac{1}{2};~ \frac{b+2}{c+3}=\frac{2}{3};~ \frac{c+3}{d+4}=\frac{3}{4};~\\inmultind~parte~cu~parte~obtinem~\frac{a+1}{b+2}* \frac{b+2}{c+3}*\frac{c+3}{d+4}=\frac{1}{2}*\frac{2}{3}*\frac{3}{4},~deci~\frac{a+1}{d+4}=\frac{1}{4};\\Din~\frac{a+1}{b+2}=\frac{1}{2};~b+2=2(a+1),~deci~b=2a.\\Din~\frac{b+2}{c+3}=\frac{2}{3},~\frac{2a+2}{c+3}=\frac{2}{3},~2(c+3)=3(2a+2),~2c+6=6a+6,~2c=6a,~deci~c=3a.\\Din~\frac{a+1}{d+4}=\frac{1}{4},~d+4=4(a+1),~d+4=4a+4,~d=4a.\\[/tex]

Explicație pas cu pas:

inlocuim in a²+b²+c²+d²=270, inlocuim, a²+(2a)²+(3a)²+(4a)²=270, ⇒a²+4·a²+9·a²+16·a²=270, ⇒a²·(1+4+9+16)=270, ⇒a²·30=270, ⇒a²=270:30=9, deci a=3.

Atunci b=2·3=6; c=3·3=9;  d=4·3=12.

Raspuns: 3,6,9,12.

[tex]\it \dfrac{a+1}{b+2}=\dfrac{1}{2} \Rightarrow 2a+2=b+2 \Rightarrow 2a=b \Rightarrow b=2a\ \ \ \ \ (1)\\ \\ \dfrac{b+2}{c+3}=\dfrac{2}{3} \Rightarrow 3b+6=2c+6 \Rightarrow 3b=2c \stackrel{(1)}{\Longrightarrow} 3\cdot2a=2c|_{:2}\Rightarrow c=3a\ \ \ \ \ (2)\\ \\ \dfrac{c+3}{d+4}=\dfrac{3}{4} \Rightarrow 4c+12=3d+12 \Rightarrow 4c=3d \stackrel{(2)}{\Longrightarrow} 4\cdot 3a = 3d|_{:3}\Rightarrow d=4a\ \ \ \ \ (3)[/tex]

Folosind relațiile (1),  (2) și (3),  ultima relație din enunț devine:

[tex]\it a^2+4a^2+9a^2+16a^2=270 \Rightarrow 30a^2=270|_{:30} \Rightarrow a^2=9 \Rightarrow a=3\ \ \ \ (4)\\ \\ (1),\ (4) \Rightarrow b=2\cdot3 \Rightarrow b=6\\ \\ (2),\ (4) \Rightarrow c=3\cdot3 \Rightarrow c=9\\ \\ (3),\ (4) \Rightarrow d=4\cdot3 \Rightarrow d=12[/tex]