\(\hept{\begin{cases}\sqrt[3]{3}=a\\\sqrt[3]{4}=b\end{cases}}\)

\(\Rightarrow b^3-a^3=1\)

\(\Leftrightarrow-b^2-ab=a^2+\frac{1}{a-b}\)

Ta cần trục cái:

\(\frac{1}{a^2-ab-b^2}=\frac{1}{a^2+a^2+\frac{1}{a-b}}=\frac{a-b}{2a^3-2a^2b+1}\)

\(=\frac{\sqrt[3]{3}-\sqrt[3]{4}}{7-2\sqrt[3]{36}}=\frac{\left(\sqrt[3]{3}-\sqrt[3]{4}\right)\left(49+14\sqrt[3]{36}+24\sqrt[3]{6}\right)}{55}=\frac{\sqrt[3]{3}-7\sqrt[3]{4}-4\sqrt[3]{18}}{55}\)

a; \(=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{3+2}=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{5}\)

b; tương tự 

\(\dfrac{1}{\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9}}=\dfrac{1}{\left(\sqrt[3]{4}\right)^2+\sqrt[3]{4}.\sqrt[3]{3}+\left(\sqrt[3]{3}\right)^2}\)

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

\(=\dfrac{\sqrt[3]{4}-\sqrt[3]{3}}{\left(\sqrt[3]{4}\right)^3-\left(\sqrt[3]{3}\right)^3}=\dfrac{\sqrt[3]{4}-\sqrt[3]{3}}{4-3}=\sqrt[3]{4}-\sqrt[3]{3}\)

a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)

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

\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)

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

\(=3\sqrt{3}\)

Vậy..

b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)

\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)

\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)

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

\(=\sqrt{5}-2-\sqrt{5}-2\)

\(=-4\)

Vậy..

Ta có: \(\frac{1}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}=\)\(\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{2}+\sqrt[3]{3}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{3}\right)^3}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{5}\)

Lời giải:

Ta có:

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

\(=\frac{(\sqrt{3}+\sqrt{5})(\sqrt{5}-1)(\sqrt{3}+1)}{(5-1)(3-1)}=\frac{(\sqrt{3}+\sqrt{5})(\sqrt{5}-1)(\sqrt{3}+1)}{8}\)

a) Ta có:

5√15+12√20+√5515+1220+5

=√52.15+√(12)2.20+√5=√25.15+√14.20+√5=√255+√204+√5=√5+√5+√5=(1+1+1)√5=3√5=52.15+(12)2.20+5=25.15+14.20+5=255+204+5=5+5+5=(1+1+1)5=35

b)  Ta có: 

√12+√4,5+√12,512+4,5+12,5

=√12+√92+√252=√12+√9.12+√25.12=√12+√32.12+√52.12=√12+3√12+5√12=(1+3+5).√12=9√12=91√2=9.√22=9√22=12+92+252=12+9.12+25.12=12+32.12+52.12=12+312+512=(1+3+5).12=912=912=9.22=922

c) Ta có:

√20−√45+3√18+√72=√4.5−√9.5+3√9.2+√36.2=√22.5−√32.5+3√32.2+√62.2=2√5−3√5+3.3√2+6√2=2√5−3√5+9√2+6√2=(2√5−3√5)+(9√2+6√2)=(2−3)√5+(9+6)√2=−√5+15√2=15√2−√520−45+318+72=4.5−9.5+39.2+36.2=22.5−32.5+332.2+62.2=25−35+3.32+62=25−35+92+62=(25−35)+(92+62)=(2−3)5+(9+6)2=−5+152=152−5

d) Ta có:

0,1√200+2√0,08+0,4.√50=0,1√100.2+2√0,04.2+0,4√25.2=0,1√102.2+2√0,22.2+0,4√52.2=0,1.10√2+2.0,2√2+0,4.5√2=1√2+0,4√2+2√2=(1+0,4+2)√2=3,4√2

Bạn giải bài đâu vậy? Kiếm điểm hỏi đáp hở, Boy anime?

1) \(=\frac{2\sqrt{3}}{\sqrt{20}}+\frac{1}{\sqrt{60}}-\frac{1}{\sqrt{15}}=\frac{6\sqrt{60}+\sqrt{60}-4\sqrt{15}}{60}=\frac{\sqrt{15}\left(12+2-4\right)}{60}=\frac{\sqrt{15}}{6}\)

a) \(=\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}\)

b) \(=\frac{12\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\frac{36+12\sqrt{3}}{9-3}=6+2\sqrt{3}\)

c) \(=\frac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{2+2\sqrt{2}+1}{2-1}=3+2\sqrt{2}\)

d) \(=\frac{\left(7\sqrt{3}-5\sqrt{11}\right)\left(8\sqrt{3}+7\sqrt{11}\right)}{\left(8\sqrt{3}-7\sqrt{11}\right)\left(8\sqrt{3}+7\sqrt{11}\right)}=\frac{217-9\sqrt{11}}{347}\)

e) \(=\frac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\frac{1+\sqrt{a}-a\sqrt{a}-a^2}{1-a}=a+\sqrt{a}+1\)

f) \(=\frac{1}{3\sqrt{2}-2\sqrt{2}+\sqrt{8}}=\frac{\sqrt{2}-\sqrt{8}}{\left(\sqrt{2}+\sqrt{8}\right)\left(\sqrt{2}-\sqrt{8}\right)}=\frac{\sqrt{2}}{6}\)

g) \(=\frac{1-\sqrt{2}+\sqrt{3}}{1-\left(\sqrt{2}-\sqrt{3}\right)^2}=\frac{1-\sqrt{2}+\sqrt{3}}{2\sqrt{6}-4}=\frac{\left(1-\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{6}+4\right)}{\left(2\sqrt{6}-4\right)\left(2\sqrt{6}+4\right)}\)

\(=\frac{2\sqrt{6}+4-4\sqrt{3}-4\sqrt{2}+6\sqrt{2}+4\sqrt{3}}{24-16}=\frac{\sqrt{2}+\sqrt{6}+2}{4}\)

f) \(=\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2\sqrt{15}-6}\)

\(=\frac{\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{15}+6\right)}{\left(2\sqrt{15}-6\right)\left(2\sqrt{15}+6\right)}=\frac{2\sqrt{30}+6\sqrt{2}-6\sqrt{5}-6\sqrt{3}+10\sqrt{3}+6\sqrt{5}}{60-36}\)

\(=\frac{\sqrt{30}+3\sqrt{2}+2\sqrt{3}}{12}\)