MCQ
$\frac{1}{\big(3+2\sqrt{2}\big)}=?$
- A$\frac{3-2\sqrt{2}}{17}$
- B$\frac{\big(3-2\sqrt{2}\big)}{13}$
- ✓$\big(3-2\sqrt{2}\big)$
- DNone of these.
✓
Answer
Correct option: C.
$\big(3-2\sqrt{2}\big)$
$\frac{1}{\big(3+2\sqrt{2}\big)}=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=\frac{3-2\sqrt{2}}{9-8}$
$=\big(3-2\sqrt{2}\big)$
Hence, the correct option is $(c).$
$=\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=\frac{3-2\sqrt{2}}{9-8}$
$=\big(3-2\sqrt{2}\big)$
Hence, the correct option is $(c).$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a:
In $E$ is the mid-point of median $AD$ such that $BE$ produced meets $AC$ at $F$. If $AC = 10.5\ cm$, then $AF =$
→Which of the following points does not lie in any quadrant?
In Fig. if l ||m, what is the value of x?
In quadrilateral $\text{ABCD , BM}$ and $DN$ are drawn perpendiculars to $AC$ such that $BM = DN$. If $BR =8 \ cm$. then $B D$ is
→Write the correct answer in the following:
AD is a diameter of a circle and $AB$ is a chord. If $AD = 34\ cm, AB = 30\ cm,$ the distance of $AB$ from the centre of the circle is:
→AD is a diameter of a circle and $AB$ is a chord. If $AD = 34\ cm, AB = 30\ cm,$ the distance of $AB$ from the centre of the circle is:
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is:
→The abscissa of a point is positive in the:
Write the correct answer in the following: $\sqrt{2}$ is a polynomial of degree.
→The base of a right triangle is $8\ cm$ and the hypotenuse is $10\ cm.$ Its area will be:
→