Rút gọn
M=\(\frac{\sqrt{x-2\sqrt{2}}}{\sqrt{x^2-4x\sqrt{2}+8}}\)- \(\frac{\sqrt{x+2\sqrt{2}}}{\sqrt{x^2+4x\sqrt{2}+8}}\)
K
Khách
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AH
Akai Haruma
Giáo viên
31 tháng 5 2019
Lời giải:
\(M=\frac{\sqrt{x-2\sqrt{2}}}{\sqrt{x^2-4x\sqrt{2}+8}}-\frac{\sqrt{x+2\sqrt{2}}}{\sqrt{x^2+4x\sqrt{2}+8}}\)
\(=\frac{\sqrt{x-2\sqrt{2}}}{\sqrt{(x-2\sqrt{2})^2}}-\frac{\sqrt{x+2\sqrt{2}}}{\sqrt{(x+2\sqrt{2})^2}}=\frac{\sqrt{x-2\sqrt{2}}}{x-2\sqrt{2}}-\frac{\sqrt{x+2\sqrt{2}}}{x+2\sqrt{2}}\)
\(=\frac{1}{\sqrt{x-2\sqrt{2}}}-\frac{1}{\sqrt{x+2\sqrt{2}}}\)
Thay $x=3$:
\(M=\frac{1}{\sqrt{3-2\sqrt{2}}}-\frac{1}{\sqrt{3+2\sqrt{2}}}=\frac{1}{\sqrt{2-2\sqrt{2}+1}}-\frac{1}{\sqrt{2+2\sqrt{2}+1}}\)
\(=\frac{1}{\sqrt{(\sqrt{2}-1)^2}}-\frac{1}{\sqrt{(\sqrt{2}+1)^2}}=\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}=\frac{2}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{2}{2-1}=2\)
KN
21 tháng 2 2020
\(ĐKXĐ:x\ge0\)
\(\left(\frac{2}{2-\sqrt{x}}+\frac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right)\)
\(=\frac{-2\sqrt{x}}{x-2\sqrt{x}}:\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{4-x}\)
\(=\frac{-2\sqrt{x}}{x-2\sqrt{x}}:\frac{\left(4+4\sqrt{x}+x\right)-\left(4-4\sqrt{x}+x\right)+4x}{4-x}\)
\(=\frac{-2\sqrt{x}}{x-2\sqrt{x}}:\frac{8\sqrt{x}+4x}{4-x}\)
\(=\frac{-2\sqrt{x}}{x-2\sqrt{x}}.\frac{4-x}{8\sqrt{x}+4x}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-2\right)\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-2\right).2\sqrt{x}\left(4+2\sqrt{x}\right)}\)
\(=\frac{\left(2+\sqrt{x}\right)}{\sqrt{x}\left(4+2\sqrt{x}\right)}=\frac{1}{2\sqrt{x}}\)
21 tháng 2 2020
mk ko kt lại nên sai từ dòng 2 r, bạn cộng thêm (3+căn x) vào r giải tương tự
3 tháng 9 2016
a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}=4x-\sqrt{8}+x=5x-\sqrt{8}\)
TT
0
17 tháng 5 2019
P=\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\left(\frac{\sqrt{x}-3}{2\sqrt{x}-x}\right)=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{4-x}\right).\frac{2\sqrt{x}-x}{\sqrt{x}-3}=\left[\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right].\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{\left(4x+8\sqrt{x}\right).\sqrt{x}.\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x}{\sqrt{x}-3}\)
29 tháng 6 2018
dk , x lơn hơn hoặc = 0 , x khác 4
\(\frac{\sqrt{x}}{\sqrt{x-2}}\times\frac{x-4}{2\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x+2}}\times\frac{x-4}{2\sqrt{x}}.\)
có \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)\)
\(\frac{\sqrt{x}}{\sqrt{x}-2}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
rút gọn
\(\frac{\left(\sqrt{x}+2\right)}{2}+\frac{\left(\sqrt{x}-2\right)}{2}\)
\(\frac{2\sqrt{x}}{2}\)