Giúp mk vs
rút gọn : 4+ $\sqrt {10 + 2$\sqrt {5} $} $ + $\sqrt {4 -$\sqrt {10 + 2$\sqrt {5} $} $} $
\(\dfrac{\sqrt{5}}{\sqrt{5}+3}+\dfrac{2\sqrt{5}}{\sqrt{5}-3}-2:\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{10}+1}\)
M.N GIÚP MK RÚT GỌN VS Ạ
\(=\dfrac{5-3\sqrt{5}+10+6\sqrt{5}}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}-\dfrac{2\sqrt{10}+2}{\sqrt{3}-\sqrt{2}}\\ =\dfrac{15+3\sqrt{5}}{5-9}-\left(2\sqrt{10}+2\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =-2\sqrt{30}-4\sqrt{5}-2\sqrt{3}-2\sqrt{2}-\dfrac{15+3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-16\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15-3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-19\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15}{4}\)
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}=}.\)
trình bày ngắn gọn giúp mk vs các tình yêu ơi
Đặt \(\sqrt{10+2\sqrt{5}}=t\)
\(VT=\sqrt{4+t}+\sqrt{4-t}\)
\(\Leftrightarrow VT^2=4+t+2\sqrt{\left(4+t\right)\left(4-t\right)}+4-t\)
\(=8+2\sqrt{16-t^2}=8+2\sqrt{6-2\sqrt{5}}\)
\(\Rightarrow VT=\sqrt{8+2\sqrt{6-2\sqrt{5}}}\) (không chắc nha)
tth làm chưa triệt để
\(VT=\sqrt{8+2\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{8+2\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{8+2\sqrt{5}-2}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
Rút gọn:
\(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
giúp mình nha!
lê thị thu huyền bn tham khảo nhá:
Bình phương 2 vế lên, ta được:
\(B^2=8+2\sqrt{4+\sqrt{10+2\sqrt{5}}4-\sqrt{10-2\sqrt{5}}}=8+2\sqrt{16-10+2\sqrt{5}}\)
\(B^2=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{"\sqrt{5-1"^2}}=8+2\sqrt{5-1}\)
Do B > 0 nên \(B=\sqrt{8+2"\sqrt{5-1}"}=\sqrt{6+2\sqrt{5}}+\sqrt{5}+1\)
P/s: Máy lác dấu ngoặc đơn phải dùng tạm ngoặc kép thông cảm
tính cụm căn của 2 lũy thừa rồi trừ; KQ;- 46 (NHỚ K ĐÚNG CHO MÌNH NHA THANKS!)
Rút gọn : \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Cho \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
B2 = \(4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
= \(8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}\)
= \(8+2\sqrt{5-2\sqrt{5}+1}\)
= \(8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
= \(8+2.\left(\sqrt{5}-1\right)\) (do \(\sqrt{5}>1\))
= \(6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1\)
= \(\left(\sqrt{5}+1\right)^2\)
=> B = \(\sqrt{5}+1\)
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}\right)^2+\left(\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2+2\sqrt{4+\sqrt{10+2\sqrt{5}}}\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.1+1^2}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(8+2\left|\sqrt{5}-1\right|=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.1+1^2\)
\(=\left(\sqrt{5}+1\right)^2\Rightarrow A=\sqrt{5}+1\left(A>0\right)\)
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\) rút gọn
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ A^2=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ A=\sqrt{5}+1\)
Rút gọn biểu thức : \(\sqrt{4+\sqrt{10+2\sqrt{5}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}}\)
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(\Leftrightarrow A^2=8+2\sqrt{16-10-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\sqrt{6-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\left(\sqrt{5}-1\right)\\ \Leftrightarrow A^2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ \Leftrightarrow A=\sqrt{5}+1\)
Vậy \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
1) rút gọn
A= \(3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
B= \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
C= \(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
Giúp mk vs ạ mk cần gấp
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\dfrac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)
\(=\dfrac{3}{2}\)
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(C=2-\sqrt{3}+3+\sqrt{3}=5\)
Rút gọn biểu thức sau :
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}-\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
(làm đầy đủ cho mk nha.Arigatou)
gọi biểu thức đó là A , ta có :
\(A^2=8+2\sqrt{16-2\sqrt{5}}\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
\(\Rightarrow A=1+\sqrt{5}\)
tk mình nhoa bạn
5.Rút gọn:
B= \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}\)
6.Rút gọn:
A= \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)