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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 :
\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)
\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)
\(=\sqrt{\left(\sqrt{3}-6\right)^2}-\sqrt{\left(\sqrt{3}+6\right)^2}\)
\(=\left|\sqrt{3}-6\right|-\left|\sqrt{3}+6\right|\)
\(=6-\sqrt{3}-\sqrt{3}-6\)
\(=-2\sqrt{3}\)
f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(3-4\right)\)
\(=\left(\sqrt{3}-1\right).\left(-1\right)=1-\sqrt{3}\)
b/ \(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
c/ \(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}=\sqrt{9}=3\)
d/ \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
\(\text{a) }\sqrt{16-8\sqrt{3}}=\sqrt{12+4-4\sqrt{12}}=\sqrt{\left(\sqrt{12}-2\right)^2}=\sqrt{12}-2\)
\(\text{b) }\sqrt{38+12\sqrt{2}}=\sqrt{36+2+12\sqrt{2}}=\sqrt{\left(6+\sqrt{2}\right)^2}=6+\sqrt{2}\)
\(\text{c) }\sqrt{22+12\sqrt{2}}=\sqrt{18+2+4\sqrt{18}}=\sqrt{\left(\sqrt{18}+\sqrt{2}\right)^2}=3\sqrt{2}+\sqrt{2}=4\sqrt{2}\)
\(\text{d) }\sqrt{17-12\sqrt{2}}=\sqrt{9+8-6\sqrt{8}}=\sqrt{\left(3-\sqrt{8}\right)^2}=3-\sqrt{8}\)
\(\text{e) }\sqrt{20-10\sqrt{3}}=\sqrt{15+5-2\sqrt{75}}=\sqrt{\left(\sqrt{15}-\sqrt{5}\right)^2}=\sqrt{15}-\sqrt{5}\)
a)
\(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4}.\sqrt{3}-\sqrt{9}.\sqrt{3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)
\(=\sqrt{3}(2-3+1)=0\)
b)
\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}.\sqrt{63}-\sqrt{4}.\sqrt{175}+\sqrt{4}.\sqrt{252}-\sqrt{4}.\sqrt{112}\)
\(=2(\sqrt{63}-\sqrt{175}+\sqrt{252}-\sqrt{112})\)
\(=2(\sqrt{9}.\sqrt{7}-\sqrt{25}.\sqrt{7}+\sqrt{36}.\sqrt{7}-\sqrt{16}.\sqrt{7})\)
\(=2(3\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7})=2\sqrt{7}(3-5+6-4)=0\)
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\(\sqrt{3}(\sqrt{12}+\sqrt{27}-\sqrt{3})=\sqrt{36}+\sqrt{81}-\sqrt{9}\)
\(=\sqrt{6^2}+\sqrt{9^2}-\sqrt{3^2}=6+9-3=12\)
c)
\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{5}}{\sqrt{7}.\sqrt{3}+\sqrt{7}.\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{7}}\)
\(\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\frac{\sqrt{81}.\sqrt{5}+3\sqrt{9}.\sqrt{3}}{3\sqrt{3}+\sqrt{9}.\sqrt{5}}=\frac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)
\(=\frac{3(3\sqrt{5}+3\sqrt{3})}{3\sqrt{3}+3\sqrt{5}}=3\)
d)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{6}+\sqrt{9}+\sqrt{12})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{2}.\sqrt{3}+\sqrt{3}.\sqrt{3}+\sqrt{3}.\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{3}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})(1-\sqrt{3})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)
\(\sqrt{5-\sqrt{21}}=\sqrt{\frac{1}{2}}.\sqrt{10-2\sqrt{21}}=\sqrt{\frac{1}{2}}.\sqrt{3-2\sqrt{3}\sqrt{7}+7}=\sqrt{\frac{1}{2}}\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{\frac{1}{2}}.\sqrt{7}-\sqrt{\frac{1}{2}}.\sqrt{3}=\sqrt{3,5}-\sqrt{1,5}\)
\(\sqrt{7+3\sqrt{5}}=\sqrt{\frac{1}{2}\left(14+2.3\sqrt{5}\right)}=\sqrt{\frac{1}{2}\left(5+2.3\sqrt{5}+3^2\right)}=\sqrt{\frac{1}{2}\left(3+\sqrt{5}\right)^2}=\sqrt{\frac{1}{2}}\left(3+\sqrt{5}\right)=\sqrt{4,5}+\sqrt{2,5}\)
\(\sqrt{49+5\sqrt{96}}=\sqrt{49+2.2.5\sqrt{6}}=\sqrt{2^2.6+2.2.5\sqrt{6}+5^2}=\sqrt{\left(5+2\sqrt{6}\right)^2}=5+2\sqrt{6}\)
\(\sqrt{5-\sqrt{21}}=\frac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7\cdot3}+3}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)
\(\sqrt{7+3\sqrt{5}}=\frac{\sqrt{14+6\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{9+2\cdot3\sqrt{5}+4}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}}{\sqrt{2}}=\frac{3+\sqrt{5}}{\sqrt{2}}\)
\(\sqrt{49+5\sqrt{96}}=\sqrt{49+5\sqrt{4\cdot24}}=\sqrt{25+2\cdot5\sqrt{24}+24}=\sqrt{\left(5+\sqrt{24}\right)^2}=5+\sqrt{24}\)
\(\sqrt{51-7\sqrt{8}}=\sqrt{51-7\sqrt{2^2\cdot2}}=\sqrt{49-2\cdot7\sqrt{2}+2}=\sqrt{\left(7+\sqrt{2}\right)^2}=7+\sqrt{2}\)
\(\sqrt{28+5\sqrt{12}}=\sqrt{28+5\sqrt{2^2\cdot3}}=\sqrt{25+2\cdot5\sqrt{3}+3}=\sqrt{\left(5+\sqrt{3}\right)^2}=5+\sqrt{3}\)
\(\sqrt{12-3\sqrt{12}}=\sqrt{12-3\sqrt{2^2\cdot3}}=\sqrt{9-2\cdot3\sqrt{3}+3}=\sqrt{\left(3+\sqrt{3}\right)^2}=3+\sqrt{3}=\sqrt{3}\left(\sqrt{3}+1\right)\)
Chúc bạn học tốt nha
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Bấm máy tính là ra thui mà bn
a/ \(=2\sqrt{3}-3\sqrt{3}+\sqrt{3}=0\)
b/ \(=\left(2\sqrt{3}-10\sqrt{3}\right)\sqrt{3}=-24\)
c/ \(=15-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=15-6\sqrt{7}\)
d/ \(=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)=12\)
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!
Lời giải:
Đặt \(\sqrt[3]{1620+12\sqrt{17457}}=a; \sqrt[3]{1620-12\sqrt{17457}}=b\)
\(\Rightarrow \left\{\begin{matrix} a^3+b^3=3240\\ ab=\sqrt[3]{1620^2-12^2.17457}=48\end{matrix}\right.\)
Tổng cần tính: \(a+b=x\)
Theo hằng đẳng thức đáng nhớ:
\((a+b)^3=a^3+b^3+3ab(a+b)\)
\(\Leftrightarrow (a+b)^3=3240+144(a+b)\)
\(\Leftrightarrow x^3=3240+144x\)
\(\Leftrightarrow (x-18)(x^2+18x+180)=0\)
Dễ thấy \(x^2+18x+180\neq 0\) nên \(x-18=0\Rightarrow x=18\)