Tinh:B=\(\sqrt{8-2\sqrt{5}}-\sqrt{8+2\sqrt{5}}\)
lam dc thi giup minh voi nghe
\(\frac{\sqrt{13+2\sqrt{11}}+\sqrt{13-2\sqrt{11}}}{\sqrt{13+5\sqrt{5}}}-\sqrt{3-2\sqrt{2}}\)
Giup minh voi
rút gọn
A=\(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
B=\(\left(1-\sqrt{5}\right)\left(\frac{\sqrt{5}+5}{2\sqrt{5}}\right)\)
C=\(\sqrt{4+\sqrt{7}}\)+\(\sqrt{4-\sqrt{7}}\)
D=\(2\sqrt{8}-4\sqrt{\left(\sqrt{2}+1\right)^2}\)
giup minh nhe các bạn
\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(0,2\sqrt{\left(-10\right)^2\cdot3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
Rut gon giup minh. Cam on
Chứng minh :
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}+8-2\sqrt{10+2\sqrt{5}}+}2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)
\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)
\(A^2=144-80-16\sqrt{5}\)
\(A^2=64-16\sqrt{5}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)
\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=2+2.\sqrt{2}.\sqrt{10}+10\)
\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)
=> \(A=\sqrt{2}+\sqrt{10}\)
Chứng minh rằng
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath
Bạn tham khảo link này!
Chứng minh \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
Biến đổi vế trái ta có :
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
= \(\sqrt{2}\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)\)
Đặt A = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
A^2 = \(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
= 8 + \(2\sqrt{16-\left(10-2\sqrt{5}\right)}\)
= \(8+2\sqrt{16-10+2\sqrt{5}}\)
= \(8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
=> A = \(\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
=> \(\sqrt{2}A=\sqrt{2}\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}=VP\) ( ĐPCM)
bn thang tran lm sai bước đưa ra hdt :v đúng là phải 16 - ( 10 + 2can5 )
= 16 - 10 - 2can5
1,thu gon bieu thuc
a A=\(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\)
b,B=\(\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}-\sqrt{7-2\sqrt{6}}}}\)
c, C=\(\dfrac{\sqrt{c^2+2c+1}}{\left|c\right|-1}\)
2,giai cac phuong trinh
a,\(x^2-9\sqrt{x}+14=0\)
b, \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-5-x^2+6\)
GIUP MINH VOI MINH CAN GAP
Cau 1:
a: \(A=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)+2\sqrt{a}\left(\sqrt{a}-2\right)}{a-4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(a+4\sqrt{a}+4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}=\sqrt{a}+2\)
c: \(=\dfrac{\left|c+1\right|}{\left|c\right|-1}\)
TH1: c>0
\(C=\dfrac{c+1}{c-1}\)
TH2: c<0
\(C=\dfrac{\left|c+1\right|}{-\left(c+1\right)}=\pm1\)
giup dum minh voi nha cac ban :
\(\sqrt{4+\sqrt{5.\sqrt{3+5.\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
Chứng minh:
\(\sqrt{\sqrt{5}-\sqrt{8-\sqrt{81-8\sqrt{5}}}}=\sqrt{2}\)
\(\sqrt{\sqrt{5}-\sqrt{8-\sqrt{81-8\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{8-\left(4\sqrt{5}-1\right)}}\)\(=\sqrt{\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-2\right)}=\sqrt{2}\)