a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

`a)(\sqrt{14}-3\sqrt{2})^2+6\sqrt{28}`

`=14-12\sqrt{7}+18+12\sqrt{7}=32`

`b)2\sqrt{20}-3\sqrt{20}+\sqrt{125}`

`=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}`

`=3\sqrt{5}`.

a) \(\left(\sqrt{14}-3\sqrt{2}\right)^2-6\sqrt{28}\)

\(=\left(\sqrt{14}\right)^2-2\cdot\sqrt{14}\cdot3\sqrt{2}+\left(3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14-6\sqrt{28}+18+6\sqrt{28}\)

\(=14+18\)

\(=32\)

b) \(2\sqrt{20}-3\sqrt{20}+\sqrt{125}\)

\(=2\cdot2\sqrt{5}-3\cdot2\sqrt{5}+5\sqrt{5}\)

\(=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}\)

\(=3\sqrt{5}\)

Áp dụng hằng đẳng thức (a+b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab.(a +b) ta có: 

\(B^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{20+14\sqrt{2}}.\sqrt[3]{20-14\sqrt{2}}\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)\)

\(B^3=40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}.B\)

\(B^3=40+3.\sqrt[3]{400-392}.B=40+3.\sqrt[3]{8}.B=40+6B\)

=> B³ - 6B - 40 = 0

<=> B³ - 64 - 6B + 24  = 0

<=> (B - 4 ).(B² + 4B + 16) - 6.(B - 4) = 0

<=> (B - 4).(B² + 4B + 16 - 6) = 0 <=> B = 4 hoặc B² + 4B + 10 = 0

B² + 4B + 10 = 0 Vô nghiêm vì \(\Delta\) = 16 - 40 = -24 < 0

Vậy B = 4

\(A=\sqrt[3]{\left(\frac{1}{2}+\frac{1}{2}\sqrt{13}\right)^3}+\sqrt[3]{\left(\frac{1}{2}-\frac{1}{2}\sqrt{13}\right)^3}\)

\(=\frac{1}{2}+\frac{\sqrt{13}}{2}+\frac{1}{2}-\frac{\sqrt{13}}{2}=1\)

\(B=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=2+\sqrt{2}+2-\sqrt{2}=4\)

cau a,b,c thay no co chung 1 dang do la

\(\sqrt[3]{a+m}+\sqrt[3]{a-m}\)

dang nay co 2 cach

C1: nhanh kho nhin de sai

VD: cau B

\(B^3=40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\left(B\right)\)

B^3=40+3(2)(B)

B^3=40+6B

B=4

C2: hoi dai nhung de nhin

dat \(a=\sqrt[3]{20+14\sqrt{2}};b=\sqrt[3]{20-14\sqrt{2}}\)

de thay B=a+b

            ab=2

            a^3+b^3=40

suy ra B^3=a^3+b^3+3ab(a+b)

B^3=40+6B

B=4

giai tuong tu

con co cach nay nhung it su dung vi kho tim

C3: dua ve tong lap phuong

VD:cau B

 \(20+14\sqrt{2}=\left(2+\sqrt{2}\right)^3\)

\(20-14\sqrt{2}=\left(2-\sqrt{2}\right)^3\)

de thay

B=4

cau d)

dung CT nay

\(\sqrt[m]{a}=\sqrt[m\cdot n]{\left(a\right)^n}\)

ap dung vao bai

\(\sqrt[3]{2\sqrt{3}-4\sqrt{2}}=\sqrt[6]{\left(2\sqrt{3}-4\sqrt{2}\right)^2}=\sqrt[6]{44-16\sqrt{6}}\)

nhanh vao

\(\sqrt[6]{\left(44-16\sqrt{6}\right)\left(44+16\sqrt{6}\right)}=\sqrt[6]{400}=\sqrt[3]{20}\)

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

a, = \(\sqrt[3]{\left(\sqrt{2}+1\right)^3}\) -  \(\sqrt[3]{\left(\sqrt{2}+1\right)^3}\) = \(\sqrt{2}-1-\sqrt{2}-1\)=-2

b, = \(\sqrt{2}+2+2-\sqrt{2}\)=4

c, = \(2+\sqrt{3}-2+\sqrt{3}\) = 2\(\sqrt{3}\)

d, = 

\(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\Leftrightarrow A^3=4+3\sqrt[3]{-1}.A\Leftrightarrow A^3=4-3A\Leftrightarrow A^3+3A-4=0\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)(1)

Ta có \(A^2+A+4>0\)

Vậy (1)\(\Leftrightarrow A-1=0\Leftrightarrow A=1\)

Vậy A=1

\(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\Leftrightarrow B^3=5\sqrt{2}+7-5\sqrt{2}+7-3\sqrt[3]{\left(5\sqrt{2}+7\right)\left(5\sqrt{2}-7\right)}\left(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\right)\Leftrightarrow B^3=14-3\sqrt[3]{1}.B\Leftrightarrow B^3=14-3B\Leftrightarrow B^3+3B-14=0\Leftrightarrow\left(B-2\right)\left(B^2+2B+7\right)=0\left(2\right)\)

Ta lại có \(B^2+2B+7>0\)

Vậy (2)\(\Leftrightarrow B-2=0\Leftrightarrow B=2\)

Vậy B=2

\(C=\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{14\sqrt{2}-20}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4+8}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4-8}=\sqrt[3]{\left(\sqrt{2}+2\right)^2}-\sqrt[3]{\left(\sqrt{2}-2\right)}=\sqrt{2}+2-\sqrt{2}+2=4\)