a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)

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

\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)

\(=5\sqrt{6}\)

b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)

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

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

\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)

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

\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)

\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)

\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)

\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)

Lời giải:

Gọi biểu thức trên là $A$

Đặt \(\sqrt[3]{15\sqrt{3}-26}=a; \sqrt[3]{15\sqrt{3}+26}=b\). Ta có:

\(a^3-b^3=-52\)

\(ab=-1\)

\(A^3=(a-b)^3=a^3-3ab(a-b)-b^3=-52+3A\)

\(\Leftrightarrow A^3-3A+52=0\)

\(\Leftrightarrow A^2(A+4)-4A(A+4)+13(A+4)=0\)

\(\Leftrightarrow (A+4)(A^2-4A+13)=0\)

Dễ thấy $A^2-4A+13>0$ nên $A+4=0$

$\Leftrightarrow A=-4$

\(B^2=\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)+\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)-2\left(4-3\right)\sqrt{26^2-3.15^2}\)

\(B^2=\left(7-4\sqrt{3}\right).\left(26+15\sqrt{3}\right)+\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)-2\)

\(B^2+2=\left(a-b\right)\left(c+d\right)+\left(a+b\right)\left(c-d\right)=ac+ad-bc-bd+ac-ad+bc-bd=2\left(ac-bd\right)\)\(B^2+2=2.\left(7.26-4.3.15\right)=2\left(182-180\right)\Rightarrow B^2=2\)

\(B>0\Rightarrow B=\sqrt{2}\)

\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)

\(=\left|4-\sqrt{15}\right|+\left|3-\sqrt{15}\right|\)

\(=4-\sqrt{15}+\sqrt{15}-3=1\)

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)

\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)

\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))

\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)

b)

\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)

\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))

\(=2\sqrt{5}-3-2\sqrt{5}+2\\ =-1\)

\(A=\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\)

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

\(=-\sqrt{2}\)