Đặt \(\sqrt[3]{a}=x;\sqrt[3]{b}=y\)

=>\(Q=\dfrac{x^4+x^2y^2+y^4}{x^2+xy+y^2}\)

\(=\dfrac{x^4+2x^2y^2+y^4-x^2y^2}{x^2+xy+y^2}\)

\(=\dfrac{\left(x^2+y^2\right)^2-\left(xy\right)^2}{x^2+xy+y^2}=\dfrac{\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)}{x^2+xy+y^2}\)

\(=x^2-xy+y^2\)

\(=\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}\)

b: \(A=\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

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

\(\Leftrightarrow A^3=4+\sqrt{15}+4-\sqrt{15}+3\cdot A\cdot1\)

\(\Leftrightarrow A^3-3A-8=0\)

hay \(A\simeq2.49\)

a: \(B=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow B^3=5-\sqrt{17}+5+\sqrt{17}+3\cdot B\cdot2=10+6B\)

\(\Leftrightarrow B^3-6B-10=0\)

hay \(B\simeq3.05\)

a) ĐS: .

b) ĐS: Nếu thì

Nếu ab

c) ĐS:

d)

Nhận xét. Nhận thấy rằng để có nghĩa thì Do đó . Vì thế có thể phân tích tử thành nhân tử.

a) ĐS: .

b) ĐS: Nếu thì

Nếu ab

c) ĐS:

d)

Nhận xét. Nhận thấy rằng để có nghĩa thì Do đó . Vì thế có thể phân tích tử thành nhân tử.

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

b.\(ab\sqrt{1+\dfrac{1}{a^2b^2}}=\sqrt{a^2b^2\left(1+\dfrac{1}{a^2b^2}\right)}=\sqrt{a^2b^2+1}\)

c.\(\sqrt{\dfrac{a}{b^3}+\dfrac{a}{b^4}}=\sqrt{\dfrac{ab+a}{b^4}}=\dfrac{\sqrt{ab+a}}{\sqrt{b^4}}=\dfrac{\sqrt{ab+a}}{b^2}\)

d. \(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+\sqrt{ab}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a-b}=\dfrac{a\sqrt{a}+a\sqrt{b}-a\sqrt{b}-b\sqrt{a}}{a-b}=\dfrac{a\sqrt{a}-b\sqrt{a}}{a-b}=\dfrac{\sqrt{a}\left(a-b\right)}{a-b}=\sqrt{a}\)

a, \(ĐKXĐ:a;b>0;a\ne2b\\ \)

Xét: \(\dfrac{2\left(a+b\right)}{\sqrt{a^3}-2\sqrt{2b^3}}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{2\left(a+b\right)}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{a+2b+\sqrt{2ab}}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}=\dfrac{1}{\sqrt{a}-\sqrt{2b}}\)\(\dfrac{\sqrt{a^3}+2\sqrt{2b^3}}{2b+\sqrt{2ab}}-\sqrt{a}=\dfrac{\left(\sqrt{a}+\sqrt{2b}\right)\left(a-\sqrt{2ab}+2b\right)}{\sqrt{2b}\left(\sqrt{a}+\sqrt{2b}\right)}-\sqrt{a}=\dfrac{\left(\sqrt{a}-\sqrt{2b}\right)^2}{\sqrt{2b}}\)\(\Rightarrow P=\dfrac{\sqrt{a}-\sqrt{2b}}{\sqrt{2b}}=\sqrt{\dfrac{a}{2b}}-1\)

b, Tự lm nhé.

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)

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

a: Sửa đề: \(\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)

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

=-1

b: Sửa đề: \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)

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

=1

Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

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

=1