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A = \(\sqrt{3+2\sqrt{2}}-\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{3+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}+3}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2+2\sqrt{3}\left(\sqrt{2}+1\right)+3}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}+1+\sqrt{3}\right)^2}\)
= \(\left|\sqrt{2}+1\right|-\left|\sqrt{2}+\sqrt{3}+1\right|\)
= \(\sqrt{2}+1-\sqrt{2}-\sqrt{3}-1\)
= \(-\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\sqrt{6+2\sqrt{5}}}{2-\sqrt{6-2\sqrt{5}}}-\dfrac{\sqrt{6-2\sqrt{5}}}{2+\sqrt{6+2\sqrt{5}}}\)
\(=\dfrac{\sqrt{5}+1}{2-\sqrt{5}+1}-\dfrac{\sqrt{5}-1}{3+\sqrt{5}}\)
\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)
\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)
\(=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$
\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)
Do đó:
\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)
\(A=\frac{\text{TS}}{\text{MS}}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\sqrt{\dfrac{1-\sqrt{3}}{1+\sqrt{3}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}}\)
\(=\sqrt{\dfrac{1-\sqrt{3}}{1+\sqrt{3}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}}\)
\(=\sqrt{\dfrac{1-\sqrt{3}}{1+\sqrt{3}}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2-\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}-1\right).\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{\dfrac{4\sqrt{3}}{2}}=\sqrt{2\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}.\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}.\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{\sqrt{2}.\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}.\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{\sqrt{2}.\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\)
\(=\sqrt{2}-\frac{\sqrt{2}}{3+\sqrt{3}}+\sqrt{2}-\frac{\sqrt{2}}{3-\sqrt{3}}\)
\(=2\sqrt{2}-\left(\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}\right)\)
\(=2\sqrt{2}-\frac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}\)
\(=2\sqrt{2}-\frac{6\sqrt{2}}{6}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sqrt[3]{a}=x;\sqrt[3]{b}=y\)
=>\(Q=\dfrac{x^4+x^2y^2+y^4}{x^2+xy+y^2}\)
\(=\dfrac{x^4+2x^2y^2+y^4-x^2y^2}{x^2+xy+y^2}\)
\(=\dfrac{\left(x^2+y^2\right)^2-\left(xy\right)^2}{x^2+xy+y^2}=\dfrac{\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)}{x^2+xy+y^2}\)
\(=x^2-xy+y^2\)
\(=\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2+\sqrt{3}}\right)}{2-2-\sqrt{3}}+\dfrac{\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)}{2-2+\sqrt{3}}\)
\(=\dfrac{-2\sqrt{2}+\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{6}+\sqrt{6+3\sqrt{3}}}{\sqrt{3}}+\dfrac{2\sqrt{2}+\sqrt{2}\left(\sqrt{3}-1\right)-\sqrt{6}-\sqrt{6-3\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{6+3\sqrt{3}}+\sqrt{6}-\sqrt{2}-\sqrt{6}-\sqrt{6-3\sqrt{3}}}{\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}\left(2+\sqrt{6}\right)+\sqrt{3}\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{2}\right)}\right)-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-3\sqrt{2}}{4-6}\right)-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{2}.\sqrt{3}}.\dfrac{4\sqrt{3}}{-2}-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{\sqrt{2}-\sqrt{3}-1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1+\left(\sqrt{2}-\sqrt{3}-1\right)\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1+2+\sqrt{6}-\sqrt{6}-3-\sqrt{2}-\sqrt{3}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{-2}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=-\dfrac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}\)
lời giải:
nhân cả hia vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}-\sqrt{2}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{8-2\sqrt{3}}-\sqrt{8+2\sqrt{3}}\)
\(=\sqrt{1^2-2\cdot1\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{1^2+2\cdot1\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\left|1-\sqrt{3}\right|-\left|1+\sqrt{3}\right|\)
\(=\sqrt{3}-1-1-\sqrt{3}\)
\(=-2\)
\(\Rightarrow A=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
vậy \(A=-\sqrt{2}\)
các bạn xem mình làm đúng chưa nè!
mình mới làm bài kiểm tra bài này, mình làm như thế nhưng mấy bạn của mình cứ bảo mình sai.
mong các bạn đóng góp ý kiến ạ!
còn một cách khác là bình phương hai vế loại bỏ giá trị dương đi thì kết quả như trên.
\(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(A=\sqrt{\left(\sqrt{3}\right)^2-1^2}-\sqrt{\left(\sqrt{3}\right)^2+1^2}\)
\(A=\left(\sqrt{\sqrt{3}-1}\right)^2-\left(\sqrt{\sqrt{3}+1}\right)^2\)
\(A=\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\)
\(A=\sqrt{3}-1-\sqrt{3}-1\)
\(A=-2\)
vay \(A=-2\)