Taifun den 22 augusti 2018 kl. 22.57

2018-08-22T22:57:50Z

Taifun: Created page with " $\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\ x^2 + 1 & = (x - 3)^2 & & \\ ..."$

2010-11-21T09:17:00Z

Created page with "<math>\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\ x^2 + 1 & = (x - 3)^2 & & \\ ..."

Ny sida

<math>\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\
x^2 + 1 & = (x - 3)^2 & & \\
x^2 + 1 & = x^2 - 6\,x + 9 & & \qquad | \;\; - x^2 \\
1 & = - 6\,x + 9 & & \qquad | \;\; + 6\,x \\
6\,x + 1 & = 9 & & \qquad | \;\; - 1 \\
6\, x & = 8 & & \qquad | \;\; / 6 \\
x & = {8 \over 6} = {4 \over 3} \\

\end{align}</math>

Prövning:

VL: <math> \sqrt{\left({4 \over 3}\right)^2 + 1} = \sqrt{{16 \over 9} + 1} = \sqrt{{16 \over 9} + {9 \over 9}} = \sqrt{{25 \over 9}} = {5 \over 3} </math>

HL: <math> {4 \over 3} - 3 = {4 \over 3} - {9 \over 3} = - {5 \over 3} </math>

VL <math> \not= </math> HL <math> \Rightarrow\, x = {4 \over 3} </math> är en falsk rot.

@@ Rad 11: / Rad 11: @@
 Prövning:
-VL: <math> \sqrt{\left({4 \over 3}\right)^2 + 1} = \sqrt{{16 \over 9} + 1} = \sqrt{{16 \over 9} + {9 \over 9}} = \sqrt{{25 \over 9}} = {5 \over 3} </math>
+VL<span style="color:black">:</span> <math> \sqrt{\left({4 \over 3}\right)^2 + 1} = \sqrt{{16 \over 9} + 1} = \sqrt{{16 \over 9} + {9 \over 9}} = \sqrt{\frac{25}{9}} = {5 \over 3} </math>
-HL: <math> {4 \over 3} - 3 = {4 \over 3} - {9 \over 3} = - {5 \over 3} </math>
+HL<span style="color:black">:</span> <math> {4 \over 3} - 3 = {4 \over 3} - {9 \over 3} = - {5 \over 3} </math>
 VL <math> \not= </math> HL <math> \Rightarrow\, x = {4 \over 3} </math> är en falsk rot.

1.1 Lösning 4a - Versionshistorik

Taifun den 22 augusti 2018 kl. 22.57

Taifun: Created page with "\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\ x^2 + 1 & = (x - 3)^2 & & \\ ..."

Taifun: Created page with " $\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\ x^2 + 1 & = (x - 3)^2 & & \\ ..."$