Giải hộ mình câu này với các bạn

Bài 3: Tính
"$|a.\sqrt{55.77.35}$ $b.~\sqrt{\frac18}.\sqrt2.\sqrt{125}.\sqrt{\frac15}$
$c.~\sqrt{\sqrt2-1}.\sqrt{\sqrt2+1}$ $d.~2\sqrt2.(\sqrt3-2)+(1+2\sqrt2)^2-2\sqrt6$"

Question

Giải hộ mình câu này với các bạn

Bài 3: Tính
"$|a.\sqrt{55.77.35}$ $b.~\sqrt{\frac18}.\sqrt2.\sqrt{125}.\sqrt{\frac15}$ 
 $c.~\sqrt{\sqrt2-1}.\sqrt{\sqrt2+1}$ $d.~2\sqrt2.(\sqrt3-2)+(1+2\sqrt2)^2-2\sqrt6$"

Ninh Hoàng · Accepted Answer

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a.

$$\sqrt{55.77.35}$$

$$=\sqrt{5.11.7.11.7.5}$$

$$=\sqrt{\left(5.7.11\right)^2}$$

$$=385$$

b.

$$\sqrt{\frac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\frac{1}{5}}$$

$$=\sqrt{\frac{1}{8}.2.125.\frac{1}{5}}$$

$$=\sqrt{\frac{1}{4}.25}$$

$$=\frac{1}{2}.5$$

$$=\frac{5}{2}$$

c.

$$\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}$$

$$=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}$$

$$=\sqrt{2-1}$$

$$=\sqrt{1}$$

$$=1$$

d.

$$2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}$$

$$=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}$$

$$=9$$.

Timi · Answer

Bài 3:
a. Ta có:
$a=\sqrt{55}	imes \sqrt{77}	imes \sqrt{35}=\sqrt{5	imes 11}	imes \sqrt{7	imes 11}	imes \sqrt{5	imes 7}$
$=\sqrt{5}	imes \sqrt{11}	imes \sqrt{7}	imes \sqrt{11}	imes \sqrt{5}	imes \sqrt{7}=5	imes 7	imes 11=385$
b. Ta có:
$b=\sqrt{\frac{1}{8}}	imes \sqrt{2}	imes \sqrt{125}	imes \sqrt{\frac{1}{5}}=\sqrt{\frac{1}{8}	imes 2}	imes \sqrt{125	imes \frac{1}{5}}$
$=\sqrt{\frac{1}{4}}	imes \sqrt{25}=\frac{1}{2}	imes 5=\frac{5}{2}$
c. Ta có:
$c=\sqrt{\sqrt{2}-1}	imes \sqrt{\sqrt{2}+1}=\sqrt{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{(\sqrt{2})^{2}-1^{2}}=\sqrt{2-1}=1$
d. Ta có:
$d=2\sqrt{2}(\sqrt{3}-2)+(1+2\sqrt{2})^{2}-2\sqrt{6}$
$=2\sqrt{2}	imes \sqrt{3}-2\sqrt{2}	imes 2+1^{2}+2	imes 1	imes 2\sqrt{2}+(2\sqrt{2})^{2}-2\sqrt{6}$
$=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}=9$