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a, \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)
Bài 2:
a: =>25x=35^2=1225
=>x=49
b: \(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
=>x+5=4
=>x=-1
1) \(A=\sqrt{x-2013}+\sqrt{2014-x}\)
Biểu thức A có nghĩa khi 2013 < hoặc = x, x < hoặc = 2014
2) \(A=\sqrt{20}+2\sqrt{80}-3\sqrt{45}\\ A=2\sqrt{5}+8\sqrt{5}-9\sqrt{5}\\ A=\sqrt{5}\left(2+8-9\right)\\ A=\sqrt{5}\)
\(2\sqrt{80}+3\sqrt{45}-\sqrt{5}\)
\(=2\cdot4\sqrt{5}+3\cdot3\sqrt{5}-\sqrt{5}\)
\(=\sqrt{5}\cdot\left(8+9-1\right)\)
\(=16\sqrt{5}\)
\(A=\frac{1}{\sqrt{11-2\sqrt{30}}}-\frac{3}{\sqrt{7-2\sqrt{10}}}+\frac{4}{\sqrt{8+4\sqrt{3}}}\)
\(=\frac{1}{\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}-\frac{3}{\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}}+\frac{2}{\sqrt{4+2\sqrt{3}}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}-\frac{3}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}}+\frac{2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{1}{\sqrt{6}-\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{2}{\sqrt{3}+1}\)
\(=\frac{6-5}{\sqrt{6}-\sqrt{5}}-\frac{5-2}{\sqrt{5}-\sqrt{2}}+\frac{3-1}{\sqrt{3}+1}\)
\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}{\sqrt{6}-\sqrt{5}}-\frac{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}+\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{3}+1}\)
\(=\sqrt{6}+\sqrt{5}-\sqrt{5}+\sqrt{2}+\sqrt{3}+1=\sqrt{6}+\sqrt{2}+\sqrt{3}+1\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)+\sqrt{3}+1=\left(\sqrt{3}+1\right)\left(\sqrt{2}+1\right)\)
\(C=\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{\frac{4-2\sqrt{3}}{2}}.\left[\sqrt{2}.\left(\sqrt{3}+\sqrt{1}\right)\right]\)
\(=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}.\sqrt{2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(D=\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{\left(8+2\sqrt{2}\right).\left(3+\sqrt{2}\right)}{9-2}-\frac{\sqrt{2}.\left(2+3\sqrt{2}\right)}{2}+\frac{\sqrt{2}.\left(1+\sqrt{2}\right)}{1-2}\)
\(=\frac{24+14\sqrt{2}+4}{7}-\frac{2\sqrt{2}+6}{2}-\frac{\sqrt{2}+2}{1}\)
\(=\frac{28+14\sqrt{2}}{7}-\sqrt{2}-3-\sqrt{2}-2\)
\(=4+2\sqrt{2}-2\sqrt{2}-5\)
\(=-1\)
\(A=\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(ĐKXĐ:x\ge0;x\ne1\right)\)
\(< =>A=\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{1}{x-\sqrt{x}}+\sqrt{x}\)
\(< =>A=\frac{1+\sqrt{x}\left(x-\sqrt{x}\right)}{x-\sqrt{x}}=\frac{1+x\sqrt{x}-x}{x-\sqrt{x}}\)
Với \(x=\frac{18}{4+\sqrt{7}}\)thì \(A=\frac{1+\frac{18}{4+\sqrt{7}}.\sqrt{\frac{18}{4+\sqrt{7}}}-\frac{18}{4+\sqrt{7}}}{\frac{18}{4+\sqrt{7}}-\sqrt{\frac{18}{4+\sqrt{7}}}}\)
\(=\frac{1}{18+\frac{4}{7}-\sqrt{18+\frac{4}{7}}}+\sqrt{18+4\sqrt{7}}\)
Em mới lớp 7 nên chỉ làm được thế thôi ạ :3
\(=3\sqrt{5}+7\sqrt{5}-4\sqrt{5}=6\sqrt{5}\)