Question

move each expression into the column with the correct equivalent expression. all expressions will be used.
drag & drop the answer
√40 - 2√10 + 4√9
√216 + 5√6
√27 + √75
√81 - √124 + 2√31 + 2√9
2√3 + 6√3
2√24 + √96 + 3√6
12
11√6
8√3
15

Explanation:

Explicación:

Paso 1: Simplificar $\sqrt{40}-2\sqrt{10}+4\sqrt{9}$

Simplificamos las raíces cuadradas: $\sqrt{40}=\sqrt{4\times10}=2\sqrt{10}$, $\sqrt{9} = 3$. Entonces $\sqrt{40}-2\sqrt{10}+4\sqrt{9}=2\sqrt{10}-2\sqrt{10}+4\times3=12$.

Paso 2: Simplificar $\sqrt{216}+5\sqrt{6}$

$\sqrt{216}=\sqrt{36\times6}=6\sqrt{6}$, entonces $\sqrt{216}+5\sqrt{6}=6\sqrt{6}+5\sqrt{6}=11\sqrt{6}$.

Paso 3: Simplificar $\sqrt{27}+\sqrt{75}$

$\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}$, $\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}$, entonces $\sqrt{27}+\sqrt{75}=3\sqrt{3}+5\sqrt{3}=8\sqrt{3}$.

Paso 4: Simplificar $\sqrt{81}-\sqrt{124}+2\sqrt{31}+2\sqrt{9}$

$\sqrt{81} = 9$, $\sqrt{124}=\sqrt{4\times31}=2\sqrt{31}$, $\sqrt{9}=3$. Entonces $\sqrt{81}-\sqrt{124}+2\sqrt{31}+2\sqrt{9}=9 - 2\sqrt{31}+2\sqrt{31}+2\times3=9 + 6=15$.

Paso 5: Simplificar $2\sqrt{3}+6\sqrt{3}$

Sumamos términos con la misma raíz: $2\sqrt{3}+6\sqrt{3}=(2 + 6)\sqrt{3}=8\sqrt{3}$.

Paso 6: Simplificar $2\sqrt{24}+\sqrt{96}+3\sqrt{6}$

$\sqrt{24}=\sqrt{4\times6}=2\sqrt{6}$, $\sqrt{96}=\sqrt{16\times6}=4\sqrt{6}$. Entonces $2\sqrt{24}+\sqrt{96}+3\sqrt{6}=2\times2\sqrt{6}+4\sqrt{6}+3\sqrt{6}=4\sqrt{6}+4\sqrt{6}+3\sqrt{6}=11\sqrt{6}$.

Respuesta:

• 12: $\sqrt{40}-2\sqrt{10}+4\sqrt{9}$
• $11\sqrt{6}$: $\sqrt{216}+5\sqrt{6}$, $2\sqrt{24}+\sqrt{96}+3\sqrt{6}$
• $8\sqrt{3}$: $\sqrt{27}+\sqrt{75}$, $2\sqrt{3}+6\sqrt{3}$
• 15: $\sqrt{81}-\sqrt{124}+2\sqrt{31}+2\sqrt{9}$

Answer:

Explicación:

Paso 1: Simplificar $\sqrt{40}-2\sqrt{10}+4\sqrt{9}$

Simplificamos las raíces cuadradas: $\sqrt{40}=\sqrt{4\times10}=2\sqrt{10}$, $\sqrt{9} = 3$. Entonces $\sqrt{40}-2\sqrt{10}+4\sqrt{9}=2\sqrt{10}-2\sqrt{10}+4\times3=12$.

Paso 2: Simplificar $\sqrt{216}+5\sqrt{6}$

$\sqrt{216}=\sqrt{36\times6}=6\sqrt{6}$, entonces $\sqrt{216}+5\sqrt{6}=6\sqrt{6}+5\sqrt{6}=11\sqrt{6}$.

Paso 3: Simplificar $\sqrt{27}+\sqrt{75}$

$\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}$, $\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}$, entonces $\sqrt{27}+\sqrt{75}=3\sqrt{3}+5\sqrt{3}=8\sqrt{3}$.

Paso 4: Simplificar $\sqrt{81}-\sqrt{124}+2\sqrt{31}+2\sqrt{9}$

$\sqrt{81} = 9$, $\sqrt{124}=\sqrt{4\times31}=2\sqrt{31}$, $\sqrt{9}=3$. Entonces $\sqrt{81}-\sqrt{124}+2\sqrt{31}+2\sqrt{9}=9 - 2\sqrt{31}+2\sqrt{31}+2\times3=9 + 6=15$.

Paso 5: Simplificar $2\sqrt{3}+6\sqrt{3}$

Sumamos términos con la misma raíz: $2\sqrt{3}+6\sqrt{3}=(2 + 6)\sqrt{3}=8\sqrt{3}$.

Paso 6: Simplificar $2\sqrt{24}+\sqrt{96}+3\sqrt{6}$

$\sqrt{24}=\sqrt{4\times6}=2\sqrt{6}$, $\sqrt{96}=\sqrt{16\times6}=4\sqrt{6}$. Entonces $2\sqrt{24}+\sqrt{96}+3\sqrt{6}=2\times2\sqrt{6}+4\sqrt{6}+3\sqrt{6}=4\sqrt{6}+4\sqrt{6}+3\sqrt{6}=11\sqrt{6}$.

Respuesta:

• 12: $\sqrt{40}-2\sqrt{10}+4\sqrt{9}$
• $11\sqrt{6}$: $\sqrt{216}+5\sqrt{6}$, $2\sqrt{24}+\sqrt{96}+3\sqrt{6}$
• $8\sqrt{3}$: $\sqrt{27}+\sqrt{75}$, $2\sqrt{3}+6\sqrt{3}$
• 15: $\sqrt{81}-\sqrt{124}+2\sqrt{31}+2\sqrt{9}$