Răspuns :
[tex]1)x \sqrt{3} - 1 = 5[/tex]
[tex]x \sqrt{3} = 5 + 1[/tex]
[tex]x \sqrt{3} = 6[/tex]
[tex]x = \frac{6}{ \sqrt{3} } [/tex]
[tex]x = \frac{6 \sqrt{3} }{3} [/tex]
[tex]x = 2 \sqrt{3} [/tex]
2)I) Metoda substituției :
[tex]\left\{\begin{matrix}
x + y = 3\\
3x - 2y = 4\end{matrix}\right.[/tex]
[tex]x + y = 3[/tex]
[tex]x = 3 - y[/tex]
[tex]3x - 2y = 4[/tex]
[tex]3(3 - y) - 2y = 4[/tex]
[tex]9 - 3y - 2y= 4[/tex]
[tex] - 3y - 2y = 4 - 9[/tex]
[tex] - 5y = - 5 \: | \div ( - 5)[/tex]
[tex]y = 1[/tex]
[tex]x = 3 - y[/tex]
[tex]x = 3 - 1[/tex]
[tex]x = 2[/tex]
II) Metoda reducerii :
[tex]\left\{\begin{matrix}
x + y = 3 \: | \times 2 \\
3x - 2y = 4\end{matrix}\right.[/tex]
[tex]\left\{\begin{matrix}
2x + 2y = 6\\
3x - 2y = 4\end{matrix}\right.[/tex]
[tex]2x + 3x + 2y - 2y = 6 + 4[/tex]
[tex]5x = 10 \: | \div 5[/tex]
[tex]x = 2[/tex]
[tex]x + y = 3[/tex]
[tex]2 + y = 3[/tex]
[tex]y = 3 - 2[/tex]
[tex]y = 1[/tex]
[tex]x \sqrt{3} = 5 + 1[/tex]
[tex]x \sqrt{3} = 6[/tex]
[tex]x = \frac{6}{ \sqrt{3} } [/tex]
[tex]x = \frac{6 \sqrt{3} }{3} [/tex]
[tex]x = 2 \sqrt{3} [/tex]
2)I) Metoda substituției :
[tex]\left\{\begin{matrix}
x + y = 3\\
3x - 2y = 4\end{matrix}\right.[/tex]
[tex]x + y = 3[/tex]
[tex]x = 3 - y[/tex]
[tex]3x - 2y = 4[/tex]
[tex]3(3 - y) - 2y = 4[/tex]
[tex]9 - 3y - 2y= 4[/tex]
[tex] - 3y - 2y = 4 - 9[/tex]
[tex] - 5y = - 5 \: | \div ( - 5)[/tex]
[tex]y = 1[/tex]
[tex]x = 3 - y[/tex]
[tex]x = 3 - 1[/tex]
[tex]x = 2[/tex]
II) Metoda reducerii :
[tex]\left\{\begin{matrix}
x + y = 3 \: | \times 2 \\
3x - 2y = 4\end{matrix}\right.[/tex]
[tex]\left\{\begin{matrix}
2x + 2y = 6\\
3x - 2y = 4\end{matrix}\right.[/tex]
[tex]2x + 3x + 2y - 2y = 6 + 4[/tex]
[tex]5x = 10 \: | \div 5[/tex]
[tex]x = 2[/tex]
[tex]x + y = 3[/tex]
[tex]2 + y = 3[/tex]
[tex]y = 3 - 2[/tex]
[tex]y = 1[/tex]
a) x√3-1=5
x√3=5+1
x√3=6
x=6/√3
x=6√3/3
x=2√3
b) x+y=3 ⇒ x=3-y
3(3-y)-2y=4
9-3y-2y=4
9-5y=4
-5y= 4-9
y=5/5
y=1
x=3-1
x=2
S(2; 1)
x√3=5+1
x√3=6
x=6/√3
x=6√3/3
x=2√3
b) x+y=3 ⇒ x=3-y
3(3-y)-2y=4
9-3y-2y=4
9-5y=4
-5y= 4-9
y=5/5
y=1
x=3-1
x=2
S(2; 1)