K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
16 tháng 7 2020
Bài 1:
a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)
\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)
\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)
\(=\sqrt{45}-1-3+\sqrt{20}\)
\(=\sqrt{45}+\sqrt{20}-4\)
\(=\sqrt{5}\left(3+2\right)-4=5\sqrt{5}-4\)
b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)
\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)
\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)
c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(\sqrt{3}+2\right)\)
\(=3\sqrt{3}+6-\sqrt{6}-2\sqrt{2}\)
d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{3}+3}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(\sqrt{7}+\sqrt{3}\right)\)
\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)
AO
2
31 tháng 8 2022
a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
24 tháng 8 2020
a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))
\(=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(3-4\right)\)
\(=-2\)
b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=8-2\sqrt{15}\)
d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)
\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|\)
\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))
\(=-3-\sqrt{6}\)
e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))
\(=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}\)
\(=\frac{16-12}{2}=\frac{4}{2}=2\)
f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=\sqrt{9}=3\)
2 tháng 2 2022
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
T
27 tháng 7 2017
a) \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(\Leftrightarrow-\left(\sqrt{3}+11\sqrt{5}+\sqrt{29}\right)\)
\(\Leftrightarrow\sqrt{637+22\sqrt{145}+2\sqrt{6\left(317+11\sqrt{145}\right)}}\)
\(\Leftrightarrow\sqrt{3}-11\sqrt{5}-\sqrt{29}\)
b) Câu hỏi của Nguyễn Trung Anh - Toán lớp 9 - Học toán với OnlineMath giống câu này!
27 tháng 7 2017
a/ \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\sqrt{5}+1=1\)
b/ Câu hỏi của Nguyễn Trung Anh - Toán lớp 9 - Học toán với OnlineMath giống câu này.
HS
11 tháng 9 2018
\(A=\dfrac{5.\left(38^2-17^2\right)}{8\left(47^2-19^2\right)}\\ =\dfrac{5\left(38-17\right)\left(38+17\right)}{8\left(47-19\right)\left(47+19\right)}\\ =\dfrac{5.21.55}{8.28.66}\\ =\dfrac{5.1155}{8.1848}\\ =\dfrac{5.5}{8.8}\\ =\dfrac{25}{64}\)
\(B=\sqrt{\dfrac{0,2\times1,21\times0,3}{7,5\times3,2\times0,64}}\\ =\sqrt{0,0625\times1,890625\times0,04}\\ =\sqrt{\dfrac{121}{25600}}\\ =\dfrac{11}{160}\)
1 tháng 7 2016
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
12 tháng 2 2022
\(x=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+3-\sqrt{5}}=\dfrac{3}{3}=1\)
\(A=\left(3\cdot1+8\cdot1+2\right)^{2018}=13^{2018}\)
\(=\left(3\sqrt{2}+\sqrt{10}\right)\left(3\sqrt{2}-\sqrt{10}\right)=18-10=8\)
sai rui anh oi :<