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a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
2)
\(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}\)
\(=\sqrt{12,1.36.10}\)
= \(\sqrt{121.36}\)
\(=\sqrt{4356}\)
\(=66\)
3)
\(\sqrt{5a}.\sqrt{45a}-3a\)
\(=\sqrt{5.45a^2}-3a\)
\(=\sqrt{225a^2}-3a\)
\(=\sqrt{\left(15a\right)^2}-3a\)
\(=-15a-3a\) ( vì \(a\le0\))
\(=-18a\)
5)
\(\sqrt{0,36a^2}\)
\(=\sqrt{\left(0,6a\right)^2}\)
\(=-0,6a\) ( vì \(a< 0\) )
Để tối mình rảnh lên coi có làm tiếp được nữa hông thì mình làm ha.
Chúc bạn học tốt!
1)
\(\sqrt{3a^3}.\sqrt{12}\)
\(=\sqrt{3}.\sqrt{a^3}.\sqrt{12}\)
\(=\sqrt{3.12}.\sqrt{a^3}\)
\(=6\sqrt{a^3}\)
4)
\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)
\(=9.6a.a^2-\sqrt{0,2}.\sqrt{18}.\sqrt{10}.\sqrt{a^2}\)
\(=54a^3-\sqrt{2}.\sqrt{18}.\sqrt{a^2}\)
\(=34a^3-\sqrt{2.18}.\sqrt{a^2}\)
\(=54a^3-6\sqrt{a^2}\)
\(=54a^3-6a^2\) ( vì a<0)
6)
\(\sqrt{a^4.\left(3-a^{ }\right)^2}\)
\(=\sqrt{\left(a^2\right)^2.\left(3-a\right)^2}\)
\(=\sqrt{\left(a^2\right)^2}.\sqrt{\left(3-a\right)^2}\)
\(=\left|a^2\right|\left|3-a\right|\) ( vì a>3 => a>3 nên 3-a<0)
Mà\(\left|3-a\right|=-\left(-3-a\right)=-3+a=a-3\)
\(=a^2\left(a-3\right)\)
\(=a^3-3a^2\)
Còn lại bạn làm tương tự nha, trể quá rùi :)))))
1. c)\(\left(\sqrt{6}-2\right)\left(\sqrt{6}+7\right)\)
\(\Leftrightarrow6+7\sqrt{6}-2\sqrt{6}-14\)
\(\Leftrightarrow-8+5\sqrt{6}\)
d)\(\left(\sqrt{3}+2\right)\left(\sqrt{3}-5\right)\)
\(\Leftrightarrow3-5\sqrt{3}+2\sqrt{3}-3\)
\(\Leftrightarrow-3\sqrt{3}\)
a) \(\sqrt{3^2}-\sqrt{7^2}+\sqrt{\left(-1\right)^2}=|3|-|7|+|-1|=3-7+1=-3\)
b) \(-2\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}+\sqrt{3^2}=-2|2|+|-5|+\left|3\right|=-4+5+3=4\)
c) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2-\sqrt{2}\right|+\left|2+\sqrt{2}\right|=2-\sqrt{2}+2+\sqrt{2}=4\)
d) \(\sqrt{\left(3\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=\left|3\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3\sqrt{2}-\sqrt{2}+1=2\sqrt{2}+1\)
e) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|+\left|\sqrt{2}+1\right|=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
f) \(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|+\left|\sqrt{5}+2\right|=\sqrt{5}-2+\sqrt{5}+2=2\sqrt{5}\)
g) \(\sqrt{9-4\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9-2\sqrt{8}}+\sqrt{2-2\sqrt{2}.3+9}=\sqrt{\left(\sqrt{8}-1\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}=\sqrt{8}-1+3-\sqrt{2}=2-\sqrt{2}+\sqrt{8}\)
h) \(\sqrt{12+8\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{12+2\sqrt{4}\sqrt{8}}+\sqrt{6-2\sqrt{2}\sqrt{4}}=\sqrt{\left(\sqrt{4}+\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}=\sqrt{4}+\sqrt{8}+\sqrt{4}-\sqrt{2}\)
k) \(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{\left(\sqrt{3}+2\right)^2}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
a,\(\sqrt{4\left(a-5\right)^2}=\sqrt{4}.\sqrt{\left(a-5\right)^2}=2.\left|a-5\right|=2\left(a-5\right)\left(a\ge5\right)\)
b,\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3=-1}\)
c,Mạn phép sửa đề ,nếu ko thì kết quả ko đẹp
\(\sqrt{8+2\sqrt{15}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{5}=\sqrt{5}+\sqrt{3}-\sqrt{5}=\sqrt{3}\)
d,\(\sqrt{\left(3-2\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=2\sqrt{3}-3-3-2\sqrt{3}=-6\)
e,\(\sqrt{24\left(b-3\right)}^2=\sqrt{24^2}.\sqrt{\left(b-3\right)^2}=24.\left(3-b\right)\left(b< 3\right)\)