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bạn chỉ cần để ý tách hằng là oke
chỉ cho này \(11+6\sqrt{2}=3^3+2.3.\sqrt{2}+\sqrt{2}^2=\left(3+\sqrt{2}\right)^2\)
và cái đằng sau nữa cũng tương tự \(11-6\sqrt{2}=\left(3-\sqrt{2}\right)^2\)
biểu thức \(< =>\sqrt{4}.\left(3+\sqrt{2}\right)^2-\sqrt{9}.\left(3-\sqrt{2}\right)^2\)ok ?
câu b tự làm đi
1. \(\left(\sqrt{5}-\sqrt{6}\right)=\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{6}+\left(\sqrt{6}\right)^2=5-2\sqrt{30}+6\)
2. \(\left(\sqrt{3}-\sqrt{5}\right)^2=\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=3-2\sqrt{15}+5\)
3. \(\left(2\sqrt{2}+\sqrt{3}\right)^2=\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2=8+4\sqrt{6}+3\)
4. \(\left(\sqrt{4}-\sqrt{17}\right)^2=\left(\sqrt{4}\right)^2-2\cdot\sqrt{4}\cdot\sqrt{17}+\left(\sqrt{17}\right)^2=4-4\sqrt{47}+17\)
5. \(\sqrt{\left(\sqrt{5}-3\right)^2}=\left|\sqrt{5}-3\right|=\left|-3+\sqrt{5}\right|=3-\sqrt{5}\)
6. \(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}+\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=4\cdot5-7=13\)
7. \(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
8. \(\sqrt{\left(5+2\sqrt{6}\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}=\left|5+2\sqrt{6}\right|-\left|5-2\sqrt{6}\right|=5+2\sqrt{6}-\left(5-2\sqrt{6}\right)=4\sqrt{6}\)9. \(\sqrt{\left(\sqrt{7}-2\right)^2}+\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-2\right|+\left|\sqrt{7}+2\right|=-2+\sqrt{7}+2+\sqrt{7}=2\sqrt{7}\)
10. \(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|+\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
#em mới lớp 8 nên không chắc lắm ạ :((
\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{3}+\sqrt{2}+1\right)\left(5\sqrt{6}-4\sqrt{3}-3\sqrt{2}+5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{3}+\sqrt{2}+1\right)\left(5\sqrt{6}-5\sqrt{3}-5\sqrt{2}+5\right)\)
\(=5\left(\sqrt{6}+\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{6}-\sqrt{3}-\sqrt{2}+1\right)\)
\(=5\left[\left(\sqrt{6}+1\right)^2-\left(\sqrt{3}+\sqrt{2}\right)^2\right]\)
\(=5.\left(6+1+2\sqrt{6}-3-2\sqrt{6}\right)\)
\(=5.2=10\)
Chúc bạn học tốt và nhớ click cho mình với nhá!
L
\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right).\sqrt{2}.\sqrt{3-\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right).\sqrt{\left(\sqrt{5-1}\right)^2}\)
= \(\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)^2\)
= \(\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
= \(2.\left(3+\sqrt{5}\right).\left(3-\sqrt{5}\right)\)
= \(2.\left(9-5\right)\)
= \(2.4=8\)
Chúc bạn học tốt !!!
Cái này mình chịu nha