1) \(P=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2}\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2+\left(4-\sqrt{15}\right)}\)

\(=\sqrt{\left(10+2\sqrt{60}+6\right)\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{\left(10+4\sqrt{15}+6\right)\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{\left(16+4\sqrt{15}\right)\cdot\left(4-\sqrt{15}\right)}\)

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

\(=\sqrt{4\left(16-15\right)}\)

\(=\sqrt{4\cdot1}\)

\(=\sqrt{4}\)

\(=2\)

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

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

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

\(=\sqrt{\left(9-6\sqrt{5}+5\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)

\(=\sqrt{\left(14-6\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)

\(=\sqrt{42+14\sqrt{5}-18\sqrt{5}-30}+\sqrt{42-14\sqrt{5}+18\sqrt{5}-30}\)

\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)

\(\sqrt{3-\sqrt{5}}\sqrt{3-\sqrt{5}}\)\(\sqrt{3+\sqrt{5}}\)\(+\sqrt{3+\sqrt{5}}\sqrt{3+\sqrt{5}}\sqrt{3-\sqrt{5}}\)

=\(\sqrt{3-\sqrt{5}}\cdot\sqrt{3^2-5}+\sqrt{3+\sqrt{5}}\cdot\sqrt{3^2-5}\)=\(2\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)=\sqrt{2}\left(\sqrt{2\cdot3-2\sqrt{5}}+\sqrt{2\cdot3+2\sqrt{5}}\right)\) =\(=\sqrt{2}\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)

b tuong tu nha ban ^.^

(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.

1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)

2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)

3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)

4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)

Lời giải:
Gọi biểu thức là A

\(A=\left[3-\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}\right]\left[\frac{\sqrt{5}(\sqrt{2}+\sqrt{3})}{\sqrt{2}+\sqrt{3}}-3\right]\)

\(=[3-\frac{-\sqrt{5}(1-\sqrt{5})}{1-\sqrt{5}}](\sqrt{5}-3)=(3--\sqrt{5})(\sqrt{5}-3)=(3+\sqrt{5})(\sqrt{5}-3)=5-3^2=-4\)

Ta có: \(\left(3-\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}+\sqrt{3}}-3\right)\)

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

=5-9

=-4

Sửa đề

\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)

\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)

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

\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)

\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)

\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)

\(d,\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{4+\sqrt{5}-1}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\sqrt{24+8\sqrt{5}}\left(\sqrt{5}-1\right)\\ =\sqrt{\left(2\sqrt{5}+2\right)^2}\left(\sqrt{5}-1\right)\\ =2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)\\ =8\)