2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)

Sửa đề

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

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

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

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

=\(\left[\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}+\dfrac{2^9.2^6.3^8}{\left(3.2\right)^6.2^9}\right]=\left[\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}+\dfrac{2^6.3^8}{3^6.2^6}\right]\)

=\(\left[\dfrac{2^5}{0,4}+3^2\right]\)

=\(\left[80+9\right]=89\)

\(\left[\dfrac{\left(2.0,4\right)^5}{0,4,0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^6.2^9}\right]\div\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}\)

\(=\left[\dfrac{2^5.0.4^5}{0,4.0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^{15}}\right]\div3^5\)

\(=\left[\dfrac{2^5}{0,4}+3^2\right]\div243\)

\(=80+\left(3^5\div3^2\right)\)

\(=80+3^3\)

\(=80+27\)

\(=107\)

f) \(\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^4}=\frac{3^3.\left(0,5\right)^5}{\left[3.\left(0,5\right)\right]^4}=\frac{3^3.\left(0,5\right)^5}{3^4.\left(0,5\right)^4}=\frac{0,5}{3}=\frac{1}{6}\)

b) \(\frac{2^3+3.2^6-4^3}{2^3+3^2}=\frac{2^3.\left(1+3.2^3-2^3\right)}{2^3+3^2}=\frac{2^3.17}{17}=2^3=8\)

Các phần còn lại tương tự, bạn tự làm nhé !

(*) Lưu ý ở những bài rút gọn có chứa lũy thừa thì bạn đưa số đó về số nguyên tố rồi thực hiện như bình thường .

VD : \(4^3=\left(2^2\right)^3=2^6\) ( đưa về số nguyên tố là 2 )

\(6^3=\left(2.3\right)^3=2^3.3^3\) ( đưa về tích hai số nguyên tố )

\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)

\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)

\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)

\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)

\(=60+15\sqrt{15}-15\sqrt{15}=60\)

a. -5  ≥  -5: Đúng

b. 4.(-3) > -14: Đúng

c. 15 < - 4 2 : Sai vì  - 4 2  = 16

d. -4 +  - 8 2 ≤  (-4).(-15): Đúng

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

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

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

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

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

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

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

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

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

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

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

\(A=4-5\)

\(A=-1\)

____

\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)

\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(B=6-121\)

\(B=-115\)