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Bài 1 : \(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-4\sqrt{45}+4}+\sqrt{45+4\sqrt{45}+4}\)
\(=\sqrt{\left(\sqrt{45}-2\right)^2}+\sqrt{\left(\sqrt{45}+2\right)^2}\)
\(=\sqrt{45}-2+\sqrt{45}+2=2\sqrt{45}\)
Bài 2 : \(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{20+6\sqrt{20}+9}+\sqrt{20-6\sqrt{20}+9}\)
\(=\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\)
\(=\sqrt{20}+3+\sqrt{20}-3=2\sqrt{20}\)
Bài 3 : \(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{27-4\sqrt{27}+4}+\sqrt{27+4\sqrt{27}+4}\)
\(=\sqrt{\left(\sqrt{27}-2\right)^2}+\sqrt{\left(\sqrt{27}+2\right)^2}\)
\(=\sqrt{27}-2+\sqrt{27}+2=2\sqrt{27}\)
Chúc bạn học tốt
4 , Ta có :
\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)
\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)
\(=\sqrt{\left(\sqrt{3}-6\right)^2}-\sqrt{\left(\sqrt{3}+6\right)^2}\)
\(=\left|\sqrt{3}-6\right|-\left|\sqrt{3}+6\right|\)
\(=6-\sqrt{3}-\sqrt{3}-6\)
\(=-2\sqrt{3}\)
a) \(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}\)
\(=\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}+1\right)^2}-\sqrt{3}\)
\(=\sqrt{3}+\sqrt{2}+1-\sqrt{3}\)
\(=\sqrt{2}+1\)
b) \(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}\)
\(=\sqrt{10+2\sqrt{15}-2\sqrt{6}-2\sqrt{10}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{2}\)
a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)
\(\sqrt{6}< \sqrt{6,25}=2,5\);
\(\sqrt{12}< \sqrt{12,25}=3,5\);
\(\sqrt{20}< \sqrt{20,25}=4,5\)
=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)
Vậy P < 12
Answer:
ý a, tham khảo bài làm của @xyzquynhdi
\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{5}+2\sqrt{3}\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
A=\(\sqrt{\left(4+\sqrt{8}\right)^2}\)\(-\sqrt{\left(4-\sqrt{8}\right)^2}\)=\(4+\sqrt{8}\)\(-\left(4-\sqrt{8}\right)\)=\(2\sqrt{8}\)
Giờ mình chỉ giải đc câu a thôi để hồi nao mình rảnh giải típ cho
22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)
\(=\frac{2\sqrt{5}}{5-2}=\frac{2\sqrt{5}}{3}\)
a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)
b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)
c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)
d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)