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Tl

Đề bài nỗi ròi

#Kirito

Sorry I can't help im only grade 4

\(\frac{5\sqrt{x}-5}{x-1}=\frac{5}{\sqrt{x}+1}\Rightarrow\sqrt{x}+1\inƯ\left(1;5\right)\)ĐK : \(x\ne1\)

\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)

\(=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{6+2\sqrt{3}-2}\)

\(=\sqrt{3+2\sqrt{3}+1}\)

\(=\sqrt{3}+1\)

\(sin57^0=cos\left(90^0-57^0\right)=cos33^0\)

\(cos66^0=cos\left(90^0-66^0\right)=cos24^0\)

\(tan77^0=cot\left(90^0-77^0\right)=cot13^0\)

\(cot57^0=tan\left(90^0-57^0\right)=tan33^0\)

6565và 5536

m khác 3 nhá chứ ko phải -3 đâu bạn ạ 

Để đths trên song song <=> \(\hept{\begin{cases}m-1=3-m\\2\ne1\end{cases}}\Leftrightarrow2m=4\Leftrightarrow m=2\)( tm ) 

\(A=\left(\frac{2\sqrt{x}+1}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)ĐK : \(x\ge0;x\ne1\)

\(=\left(\frac{2\sqrt{x}+1-\left(\sqrt{x}-1\right)}{x-1}\right):\frac{1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}+2}{x-1}:\frac{1}{\sqrt{x}-1}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(B=\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1-\sqrt{x}}{\sqrt{x}-2}-\frac{\sqrt{x}+4}{x-\sqrt{x}-2}\)ĐK : \(x\ge0;x\ne4\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)+\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x-2\sqrt{x}+1-x-\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}==\frac{-3}{\sqrt{x}-2}\)

A =( \(\frac{2\sqrt{x}+1}{x-1}\)\(-\) \(\frac{1}{\sqrt{x}+1}\)) \(\div\) \(\frac{1}{\sqrt{x}-1}\)ĐK: x\(\ge0\)và x\(\ne1\)

   =(\(\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)\(-\)\(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)) \(\div\)\(\frac{1}{\sqrt{x}-1}\)

   =\(\frac{2\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)\(\times\)\(\frac{\sqrt{x}-1}{1}\)

   =\(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)\(\times\)\(\frac{\sqrt{x}-1}{1}\)

   =\(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

Vậy A = \(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)với x\(\ge0\)và x\(\ne1\)