\(x^2=\left(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\right)^2\)

\(x^2=21+6\sqrt{6}+21-6\sqrt{6}-2\sqrt{441-216}\)

\(x^2=42-2\sqrt{225}\)

\(x^2=42-30=12\)

\(x=2\sqrt{3}\)

nếu có sai bn thông cảm nha

cách khác nhé:

\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)

\(=\sqrt{21+2.3\sqrt{2}.\sqrt{3}}-\sqrt{21-2.3\sqrt{2}.\sqrt{3}}\)

\(=\sqrt{18+2.\sqrt{18}.\sqrt{3}+3}-\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}\)

\(=\left(\sqrt{18}+\sqrt{3}\right)-\left(\sqrt{18}-\sqrt{3}\right)\)

\(=2\sqrt{3}\)

p/s: mk đã phân tích kĩ ra cho bn rồi đó

\(A=\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}\right)^2-2.3.\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(A=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(A=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)

Cảm ơn ạ

Lê Hùng Hải không có gì :))

tuổi con HN là :

50 : ( 1 + 4 ) = 10 ( tuổi )

tuổi bố HN là :

50 - 10 = 40 ( tuổi )

hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi

ta có sơ đồ : bố : |----|----|----|

                  con : |----| hiệu 30 tuổi

tuổi con khi đó là :

 30 : ( 3 - 1 ) = 15 ( tuổi )

số năm mà bố gấp 3 tuổi con là :

 15 - 10 = 5 ( năm )

       ĐS : 5 năm

mình nha

bạn làm bài nào thế ?

bn làm bài như thế nào z

hok tốt

\(\sqrt{3+\sqrt{5}}=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}+1}{\sqrt{2}}\)

\(\sqrt{7+3\sqrt{5}}=\frac{\sqrt{14+2.3\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{9+2.3\sqrt{5}+5}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}}{\sqrt{2}}=\frac{3+\sqrt{5}}{\sqrt{2}}\)

\(\sqrt{21+6\sqrt{6}}=\sqrt{3+2.\sqrt{3}.3\sqrt{2}+18}=\sqrt{\left(\sqrt{3}+3\sqrt{2}\right)^2}=\sqrt{3}+3\sqrt{2}\)

\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=3\sqrt{2}-\sqrt{3}\)

Nên \(E=\frac{\sqrt{5}+1+3+\sqrt{5}}{\sqrt{2}}.\left(3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\right)\)

\(=\frac{4+2\sqrt{5}}{\sqrt{2}}.2.3.\sqrt{2}=24+12\sqrt{5}\)

Cung Bảo Bình rất uy tín

Lời giải:

Đặt \(\sqrt[3]{27+6\sqrt{21}}=a; \sqrt[3]{27-6\sqrt{21}}=b\) thì ta cần tính tổng $A=a+b$.

Ta có:

$a^3+b^3=54$

\(ab=\sqrt[3]{(27+6\sqrt{21})(27-6\sqrt{21})}=-3\)

$A^3=(a+b)^3=a^3+b^3=3ab(a+b)=54+3(-3)A$

$\Leftrightarrow A^3=54-9A$

$\Leftrightarrow A^3+9A-54=0$

$\Leftrightarrow A^2(A-3)+3A(A-3)+18(A-3)=0$

$\Leftrightarrow (A^2+3A+18)(A-3)=0$

$\Leftrightarrow A-3=0$ (do $A^2+3A+18>0$)

$\Leftrightarrow A=3$

\(a.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.\sqrt{18}.\sqrt{3}+3}+\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}=\sqrt{18}+\sqrt{3}+\sqrt{18}-\sqrt{3}=2\sqrt{18}=6\sqrt{2}\)

1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)

3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)

5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)

\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)

\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)

\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)

\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)

a)

=\(\sqrt{18-2.3\sqrt{2}.1+1}\)

\(=\sqrt{\left(3\sqrt{2}-1\right)^2}\)

\(=3\sqrt{2}-1\)

b)

=\(\sqrt{12+2.2\sqrt{3}.3+9}\)

=\(\sqrt{\left(2\sqrt{3}+3\right)^2}\)

=\(2\sqrt{3}+3\)

c)

=\(\sqrt{25-2.5.4\sqrt{2}+32}\)

=\(\sqrt{\left(5-4\sqrt{2}\right)^2}\)

=\(4\sqrt{2}-5\)

d)

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

e)

\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)

g)

\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

b, \(\sqrt{21+12\sqrt{3}}=\sqrt{21+2.3.2.\sqrt{3}}=\sqrt{9+2.3.\sqrt{12}+12}\)

\(=\sqrt{\left(3+\sqrt{12}\right)^2}=3+\sqrt{12}\)

\(c,\sqrt{57-40\sqrt{2}}=\sqrt{57-2.4.5.\sqrt{2}}=\sqrt{25-2.5.\sqrt{32}}\)

\(=\sqrt{\left(5-\sqrt{32}\right)^2}=\left|5-\sqrt{32}\right|=5-\sqrt{32}\)

\(d,\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{\left(3-2.\sqrt{2}.\sqrt{3}+2\right)\left(3-2\sqrt{3}+1\right)}\) \(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\)

\(e,A=\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)

Ta có :

\(21+6\sqrt{6}=\dfrac{42+12\sqrt{6}}{2}=\dfrac{36+2.6.\sqrt{6}+6}{2}=\left(\dfrac{6+\sqrt{6}}{\sqrt{2}}\right)^2\) Tương tự : \(21-6\sqrt{6}=\left(\dfrac{6-\sqrt{6}}{\sqrt{2}}\right)^2\)

Do đó :

\(A=\sqrt{\left(\dfrac{6+\sqrt{6}}{\sqrt{2}}\right)^2}+\sqrt{\left(\dfrac{6-\sqrt{6}}{\sqrt{2}}\right)^2}=\dfrac{6+\sqrt{6}}{\sqrt{2}}+\dfrac{6-\sqrt{6}}{\sqrt{2}}=\dfrac{6+\sqrt{6}+6-\sqrt{6}}{\sqrt{2}}\)\(=\dfrac{12}{\sqrt{2}}=\dfrac{12\sqrt{2}}{2}=6\sqrt{2}\)

Phần g làm tương tự như phần e nha bạn :>

Chúc bạn học tốt :>