Question

Help me with this question please

Answer 1

Answer:

$\dfrac{991}{40\sqrt{2}}$

Step-by-step explanation:

Given expression:

$\sqrt{72}-\dfrac{48}{\sqrt{50}}-\dfrac{45}{\sqrt{128}}+2\sqrt{98}$

Rewrite 72 as 36·2, 50 as 25·2, 128 as 64·2 and 98 as 49·2:

$\implies \sqrt{36 \cdot 2}-\dfrac{48}{\sqrt{25 \cdot 2}}-\dfrac{45}{\sqrt{64 \cdot 2}}+2\sqrt{49 \cdot 2}$

$\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:$

$\implies \sqrt{36}\sqrt{2}-\dfrac{48}{\sqrt{25}\sqrt{2}}-\dfrac{45}{\sqrt{64}\sqrt{ 2}}+2\sqrt{49}\sqrt{2}$

Rewrite 36 as 6², 25 as 5², 64 as 8² and 49 as 7²:

$\implies \sqrt{6^2}\sqrt{2}-\dfrac{48}{\sqrt{5^2}\sqrt{2}}-\dfrac{45}{\sqrt{8^2}\sqrt{ 2}}+2\sqrt{7^2}\sqrt{2}$

$\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0$

$\implies 6\sqrt{2}-\dfrac{48}{5\sqrt{2}}-\dfrac{45}{8\sqrt{ 2}}+2\cdot 7\sqrt{2}$

Simplify:

$\implies 6\sqrt{2}-\dfrac{48}{5\sqrt{2}}-\dfrac{45}{8\sqrt{ 2}}+14\sqrt{2}$

Combine like terms:

$\implies 20\sqrt{2}-\dfrac{48}{5\sqrt{2}}-\dfrac{45}{8\sqrt{ 2}}$

Make the denominators of the two fractions the same:

$\implies 20\sqrt{2}-\dfrac{384}{40\sqrt{2}}-\dfrac{225}{40\sqrt{ 2}}$

Rewrite 20√2 as a fraction with denominator 40√2:

$\implies 20\sqrt{2}\cdot\dfrac{40\sqrt{2}}{40\sqrt{2}}-\dfrac{384}{40\sqrt{2}}-\dfrac{225}{40\sqrt{ 2}}$

$\implies \dfrac{1600}{40\sqrt{2}}-\dfrac{384}{40\sqrt{2}}-\dfrac{225}{40\sqrt{ 2}}$

Combine fractions:

$\implies \dfrac{991}{40\sqrt{2}}$