Відповідь:
1) (х+2)²=5х+7
x²+4x+4=5x+7
x²+4x+4-5x-7=0
x²-x-3=0
[tex]x=\frac{-(-1)+-\sqrt{(-1)^2-4*1*(-3)}}{2*1}[/tex]
[tex]x=\frac{1+-\sqrt{1+12}}{2}[/tex]
[tex]x=\frac{1+-\sqrt{13}}{2}[/tex]
[tex]x1=\frac{1-\sqrt{13}}{2} , x2=\frac{1+\sqrt{13}}{2}[/tex]
2) 1/4х²-х-5=0
[tex]\frac{1}{4}x^{2} -x-5=0[/tex]
[tex]x^{2} -4x-20=0[/tex]
[tex]x=\frac{-(-4)+-\sqrt{(-4)^2-4*1*(-20)}}{2*1}[/tex]
[tex]x=\frac{4+-\sqrt{16+80}}{2}[/tex]
[tex]x=\frac{4+-\sqrt{96}}{2}[/tex]
[tex]x=\frac{4+-4\sqrt{6}}{2}[/tex]
{x1 = 2-2√6
{x2 = 2+2√6
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
Відповідь:
1) (х+2)²=5х+7
x²+4x+4=5x+7
x²+4x+4-5x-7=0
x²-x-3=0
[tex]x=\frac{-(-1)+-\sqrt{(-1)^2-4*1*(-3)}}{2*1}[/tex]
[tex]x=\frac{1+-\sqrt{1+12}}{2}[/tex]
[tex]x=\frac{1+-\sqrt{13}}{2}[/tex]
[tex]x1=\frac{1-\sqrt{13}}{2} , x2=\frac{1+\sqrt{13}}{2}[/tex]
2) 1/4х²-х-5=0
[tex]\frac{1}{4}x^{2} -x-5=0[/tex]
[tex]x^{2} -4x-20=0[/tex]
[tex]x=\frac{-(-4)+-\sqrt{(-4)^2-4*1*(-20)}}{2*1}[/tex]
[tex]x=\frac{4+-\sqrt{16+80}}{2}[/tex]
[tex]x=\frac{4+-\sqrt{96}}{2}[/tex]
[tex]x=\frac{4+-4\sqrt{6}}{2}[/tex]
{x1 = 2-2√6
{x2 = 2+2√6