Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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!
Bạn chỉ cần tách chúng thành hằng đẳng thức sau đó áp dụng HĐT: \(\sqrt{A^2}=\left|A\right|\)
1, \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}+1\right|=\sqrt{2}+1\)
2, \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)3, \(\sqrt{5+2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|=\sqrt{3}+\sqrt{2}\)4, \(\sqrt{7-2\sqrt{10}}=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left|\sqrt{5}-\sqrt{2}\right|=\sqrt{5}-\sqrt{2}\)5, \(\sqrt{15-6\sqrt{6}}=\sqrt{9-2.3.\sqrt{6}+6}=\sqrt{\left(3+\sqrt{6}\right)^2}=\left|3+\sqrt{6}\right|=3+\sqrt{6}\)Các câu còn lại tương tự nha!
6, \(\sqrt{8+2\sqrt{15}}=\sqrt{5+2\sqrt{5}.\sqrt{3}+3}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left|\sqrt{5}+\sqrt{3}\right|=\sqrt{5}+\sqrt{3}\)7, \(\sqrt{10-2\sqrt{21}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left|\sqrt{7}-\sqrt{3}\right|=\sqrt{7}-\sqrt{3}\)8, \(\sqrt{11+2\sqrt{18}}=\sqrt{9+2\sqrt{9}.\sqrt{2}+2}=\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}=\left|3+\sqrt{2}\right|=3+\sqrt{2}\)9, \(\sqrt{14-2\sqrt{33}}=\sqrt{11-2\sqrt{11}.\sqrt{3}+3}=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\left|\sqrt{11}-\sqrt{3}\right|=\sqrt{11}-\sqrt{3}\)Thử tự làm những câu còn lại rồi kiểm tra xem đúng hay sai nha!!!
Chúc bạn học tốt!!!
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
https://hoc24.vn/hoi-dap/question/407636.html
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}\)
= 9
~ ~ ~ ~ ~
\(M=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
1: \(=\sqrt{36}=6\)
2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)
3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)
4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)
bạn giải chi tiết giúp mk đc k ạ