Với \(x\ge0;x\ne1\)

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

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

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

1, \(\sqrt{12}+5\sqrt{3}-\sqrt{48}=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)

2, \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}=5\sqrt{5}+2\sqrt{5}-6\sqrt{5}=\sqrt{5}\)

còn đâu bạn tự làm, tương tự 

7, \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9+2.3\sqrt{2}+2}+\sqrt{9-2.3\sqrt{2}+2}\)

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

8, \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

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

9, \(\sqrt{8-2\sqrt{7}}-\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{7-4\sqrt{7}+4}\)

\(=\sqrt{7}-1-\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-1-\sqrt{7}+2=1\)

\(\frac{2}{\sqrt{3}-1}-\frac{1}{\sqrt{3}-2}+\frac{12}{\sqrt{3}+3}=\frac{2\sqrt{3}+2}{3-1}-\frac{\sqrt{3}+2}{3-2}+\frac{12\sqrt{3}-36}{3-9}\)

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

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

\(=\frac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\frac{11}{a-9}\)

Để B nguyên thì \(\frac{11}{a-9}\inℤ\Leftrightarrow a-9\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

đến đây bạn tự lập bảng xét ước nhé :c chú ý ĐK giùm mình không lại sai :>

\(\left(x^2+2x+1\right)^2+2\left(x^2+2x+1\right)+1=x\)

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

Với : x >= 0 => \(x^2+2x+2>0\)

\(\Leftrightarrow x^2+2x+2=\sqrt{x}\Leftrightarrow x^2+2x+2-\sqrt{x}=0\) 

\(\Delta=4-4\left(2-\sqrt{x}\right)=4-8+4\sqrt{x}=-4+4\sqrt{x}\)

Để pt có nghiệm khi delta >=0 

\(-4+4\sqrt{x}\ge0\Leftrightarrow1-\sqrt{x}\le0\Leftrightarrow0\le x\le1\)

\(ĐKXĐ:x\le2\)

\(x^2=\sqrt{2-x}+2\)

\(x^2-2=\sqrt{2-x}\)

\(x^4-4x^2+4=2-x\)

\(x^4-4x^2+x+2=0\)

\(x^4+x^3-3x^2-2x-1x^3-x^2+3x+2=0\)

\(x\left(x^3+x^2-3x-2\right)-\left(x^3+x^2-3x-2\right)=0\)

\(\left(x-1\right)\left(x^3+x^2-3x-2\right)=0\)

\(\left(x-1\right)\left(x^3+2x^2-x^2-2x-x-2\right)=0\)

\(\left(x-1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)-\left(x+2\right)\right]=0\)

\(\left(x-1\right)\left(x+2\right)\left(x^2-x-1\right)=0\)

\(TH1:x-1=0< =>x=1\left(KTM\right)\)

mình cũng chưa biết tại sao nó lại không thỏa mãn nữa :v

\(TH2:x+2=0< =>x=-2\left(TM\right)\)

xét \(x^2-x-1=0\)

\(\sqrt{\Delta}=\sqrt{\left(-1\right)^2-4.1.\left(-1\right)}=\sqrt{5}\)

\(\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\left(TM\right)\\x_2=\frac{1-\sqrt{5}}{2}\left(KTM\right)\end{cases}}\)

vậy kl...................

dốt quá bạn thêm đkxđ vào lúc bình cả hai vế lên nha

\(\orbr{\begin{cases}x\le-\sqrt{2}\\x\ge\sqrt{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=1\left(KTM\right)\\x=\frac{1-\sqrt{5}}{2}\left(KTM\right)\end{cases}}\)