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1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tớ ko chép lại đề, kí hiệu nhé
(1) \(=\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{\left|\sqrt{6}+\sqrt{5}\right|^2}=\left(\sqrt{6}-\sqrt{5}\right)^2-\left(\sqrt{6}+\sqrt{5}\right)=1-2\sqrt{30}-\sqrt{6}-\sqrt{5}\)
ai ra đề mà để đáp án dài thế này mất thẩm mĩ quá!!!
(2) \(=\sqrt{\left|\sqrt{5}+\sqrt{3}\right|^2}-\sqrt{\left|\sqrt{5}-\sqrt{3}\right|^2}=\left(\sqrt{5}+\sqrt{3}\right)-\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}\)
(3) \(=\sqrt{\left|\sqrt{7}+2\right|^2}-\sqrt{\left|3-\sqrt{5}\right|^2}=\sqrt{7}+2-3+\sqrt{5}=\sqrt{7}+\sqrt{5}-1\)
lại thêm 1 phép tính không đẹp....
(4) \(=\sqrt{\left|3\sqrt{2}-2\right|^2}-\sqrt{\left|3\sqrt{2}+1\right|^2}=3\sqrt{2}-2-3\sqrt{2}-1=-3\)
(5) \(=\sqrt{\left|2\sqrt{3}-1\right|^2}+\sqrt{\left|2\sqrt{3}-3\right|^2}=2\sqrt{3}-1+2\sqrt{3}-3=4\sqrt{3}-4\)
kiểm tra lại kết quả nhé ^^! Cảm ơn!
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a) \(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2.2}=\sqrt{4}=2\)
b) \(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{8+2\sqrt{15}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)^2\left(\sqrt{15}-4\right)=2\left(4+\sqrt{15}\right)\left(\sqrt{15}-4\right)\)
\(=-2\)
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Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)
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a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)
\(=\sqrt{\left(6-\sqrt{11}\right)\left(6+\sqrt{11}\right)}\)
\(=\sqrt{6^2-\left(\sqrt{11}\right)^2}\)
\(=\sqrt{36-11}\)
\(=\sqrt{25}\)
\(=\sqrt{5^2}\)
\(=5\)
b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
\(=\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}\)
\(=\sqrt{8^2-\left(\sqrt{15}\right)^2}\)
\(=\sqrt{64-15}\)
\(=\sqrt{49}\)
\(=\sqrt{7^2}\)
\(=7\)
a: \(=\sqrt{6^2-11}=\sqrt{25}=5\)
b: \(=\sqrt{8^2-15}=\sqrt{49}=7\)
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1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
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a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
mik bật mí cho
nhưng nhớ
kickkkkkkkkkkkkkk đồng ý ko
\(\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left|\sqrt{5}+\sqrt{3}\right|=\sqrt{5}+\sqrt{3}\)
\(\sqrt{11+4\sqrt{6}}=\sqrt{11+2.2\sqrt{6}}=\sqrt{\left(\sqrt{6}\right)^2+2.2\sqrt{6}+2^2}\)
\(=\sqrt{\left(\sqrt{6}+2\right)^2}=\left|\sqrt{6}+2\right|=\sqrt{6}+2\)