a) ta có 2=1+1=\(\sqrt{1}\)+1

Vì \(\sqrt{1}\)<\(\sqrt{2}\)nên \(\sqrt{1}\)+1  <   \(\sqrt{2}\)+1

 \(\Rightarrow\)2<\(\sqrt{2}\)+1

b) ta có: 12=\(\sqrt{25}\)+\(\sqrt{49}\)

\(\Rightarrow\)\(\hept{\begin{cases}\sqrt{24}< \sqrt{25}\\\sqrt{45}< \sqrt{49}\end{cases}}\Rightarrow\)\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}\)hay \(\sqrt{24}+\sqrt{45}\)<12

chịu thôi

\(x^4+2x^3+2x^2-2x+1=\left(x^3+x\right)\sqrt{\frac{1-x^2}{x}}\)ĐK: \(0< x< 1\)

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

\(\Leftrightarrow x^2\left(x+1\right)^2+\left(1-x\right)^2=\left(x^2+1\right)\sqrt{x\left(x+1\right)\left(1-x\right)}\left(1\right)\)

Đặt  \(\hept{\begin{cases}x\left(x+1\right)=a>0\\1-x=b>0\end{cases}}\)\(\Rightarrow x^2+1=a+b\)

\(\Rightarrow\left(1\right)\Leftrightarrow a^2+b^2=\left(a+b\right)\sqrt{ab}\)

\(\Leftrightarrow a^2+b^2-a\sqrt{ab}-b\sqrt{ab}=0\)

\(\Leftrightarrow a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+b\sqrt{b}\left(\sqrt{b}-\sqrt{a}\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a^3}-\sqrt{b^3}\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\left(a+\sqrt{ab}+b\right)=0\)

Vì \(a+\sqrt{ab}+b=\left(\sqrt{a}+\frac{\sqrt{b}}{2}\right)^2+\frac{3\sqrt{b}}{4}>0;\forall a,b>0\)

\(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=0\)

\(\Leftrightarrow a=b\)

\(\Rightarrow x^2+x=1-x\)

\(\Leftrightarrow x^2+2x-1=0\)

\(\Delta=8\)

=> pt có 2 nghiệm pb \(^{\orbr{\begin{cases}x=-1+\sqrt{2}\left(tm\right)\\x=-1-\sqrt{2}\left(loai\right)\end{cases}}}\)

Vậy ...

quy đồng em 

\(\frac{1}{5+3\sqrt{2}}+\frac{1}{5-3\sqrt{2}}=\frac{5-3\sqrt{2}+5+3\sqrt{2}}{25-9.2}=\frac{10}{7}\)

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

=\(\frac{5-3\sqrt{2}}{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}\)+\(\frac{5+3\sqrt{2}}{\left(5-3\sqrt{2}\right)\left(5+3\sqrt{2}\right)}\)

=\(\frac{5+3\sqrt{2}}{25-18}\)+\(\frac{5-3\sqrt{2}}{25-18}\)

=\(\frac{5+3\sqrt{2}+5-3\sqrt{2}}{25-18}\)

=\(\frac{10}{7}\)

\(B=x-2\sqrt{x-1}\)

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

\(B=\left(\sqrt{x-1}-1\right)^2\)

ta có \(B\in Z^+< =>B\ge1\)và B nguyên

\(< =>\left(\sqrt{x-1}-1\right)^2\ge1\)

\(x\ge5\)và x nguyên

Để B thuộc Z+ thì ta có :

\(\hept{\begin{cases}\sqrt{x-1}\in Z^+\\x>2\sqrt{x-1}\end{cases}\Leftrightarrow\hept{\begin{cases}x=a^2+1\text{ với }a\in N\\x^2>4x-4\Leftrightarrow x\ne2\end{cases}}}\)

Vậy x có dạng \(a^2+1\text{ với }a\in N\text{ và }a\ne1\)