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b) bạn trục mẫu đi nha dựa vào hằng đẳng thức a^2 -b^2=(a-b)(a+b)
rồi bạn tính nói chung mẫu bằng -1
tính cái trên tử kết quả là 4
c) bạn dựa vào câu b .\(\dfrac{1}{\sqrt{3}}=\dfrac{2}{2\sqrt{3}}>\dfrac{2}{\sqrt{3}+\sqrt{4}}\)
từ đó suy ra B > 2A vậy B>8
Lời giải:
a) \(\frac{1}{1-\sqrt[3]{5}}=\frac{1+\sqrt[3]{5}+\sqrt[3]{5^2}}{(1-\sqrt[3]{5})(1+\sqrt[3]{5}+\sqrt[3]{25})}\) \(=\frac{1+\sqrt[3]{5}+\sqrt[3]{25}}{1^3-5}=\frac{1+\sqrt[3]{5}+\sqrt[3]{25}}{-4}\)
b)
\(\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}}=\frac{\sqrt[3]{2^2}-\sqrt[3]{6}+\sqrt[3]{3^2}}{(\sqrt[3]{2}+\sqrt[3]{3})(\sqrt[3]{2^2}-\sqrt[3]{6}+\sqrt[3]{3^2})}\) \(=\frac{\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}}{2+3}=\frac{\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}}{5}\)
c)
\(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{2^2}+\sqrt[3]{2}+1)}=\frac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1\)
a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)
\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)
b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)
\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)
c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)
\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)
bài 2:
a: \(\dfrac{25}{5-2\sqrt{3}}=\dfrac{125+10\sqrt{3}}{13}\)
b: \(\dfrac{8}{\sqrt{5}+2}=8\sqrt{5}-32\)
c: \(\dfrac{6}{2\sqrt{3}-\sqrt{7}}=\dfrac{12\sqrt{3}+6\sqrt{7}}{5}\)
d: \(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\dfrac{\sqrt{6}}{2}\)
a)
\(\sqrt{\dfrac{27a^4}{48a^2}}=\sqrt{\dfrac{9a^2}{16}}=\sqrt{\left(\dfrac{3a}{4}\right)^2}=\dfrac{3a}{4}\)
b)
\(\dfrac{\sqrt{9x^2-25}}{\sqrt{3x+5}}=\dfrac{\sqrt{\left(3x-5\right)\left(3x+5\right)}}{\sqrt{3x+5}}=\sqrt{3x-5}\)
c)
\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\\ =\left(3-\sqrt{5}\right)+2.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\\ =2.\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+6\\ =2.\sqrt{9-5}+6\\ =10\\ \Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{10}\)
d) KO khó!!
câu e mình viết sai đề, mk sửa lại nhé , với mình bổ sung câu f
e) \(\dfrac{2}{\sqrt[3]{4}+\sqrt[3]{5}}\)
f) \(\dfrac{1}{2-\dfrac{\sqrt[3]{3}}{2}}\)
Bài 3:
a: \(=\dfrac{3+2\sqrt{2}}{1}-\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}\)
\(=3+2\sqrt{2}-\sqrt{2}=3+\sqrt{2}\)
b: \(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\left(\sqrt{ab}-b\right)}{\left(a+\sqrt{b}\right)^2}\)
\(=\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=\dfrac{b}{a+\sqrt{b}}\)
c: \(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)
bài 1) a) \(xy\sqrt{\dfrac{x}{y}}=x\sqrt{y}\sqrt{y}\dfrac{\sqrt{x}}{\sqrt{y}}=x\sqrt{x}\sqrt{y}=\left(\sqrt{x}\right)^3\sqrt{y}\)
b) \(\sqrt{\dfrac{5a^3}{49b}}=\dfrac{\sqrt{5a^3}}{\sqrt{49b}}=\dfrac{\sqrt{5a^3}}{7\sqrt{b}}=\dfrac{\sqrt{5a^3}.\sqrt{b}}{7\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{5a^3b}}{7b}\)
bài 2) a) \(\dfrac{\sqrt{3}-3}{1-\sqrt{3}}=\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=\sqrt{3}\)
b) \(\dfrac{5-\sqrt{15}}{\sqrt{3}-\sqrt{5}}=\dfrac{-\sqrt{5}\left(\sqrt{3}-\sqrt{5}\right)}{\sqrt{3}-\sqrt{5}}=-\sqrt{5}\)
c) \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
a: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
b: \(=\sqrt{2}-1-\sqrt{2}=-1\)