a, ĐKXĐ \(x\ne0,1\)

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

\(=\frac{3x}{x\left(x-1\right)}=\frac{3}{x-1}\)

b, Để B nguyên thì \(3⋮x-1\)

\(\Rightarrow x-1\in\left\{1,3,-1,-3\right\}\)

\(\Rightarrow x\in\left\{2,4,0,-2\right\}\)

\(ĐKXĐ:m\ge0;m\ne1\)

\(a,M=\frac{\sqrt{m}-1}{\sqrt{m}+1}+\frac{\sqrt{m}+1}{\sqrt{m}-1}\)

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

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

\(=\frac{2m+2}{m-1}\)

b,Để M nguyên thì \(\frac{2m+2}{m-1}=\frac{2\left(m-1\right)}{m-1}+\frac{4}{m-1}=2+\frac{4}{m-1}\) nguyên

\(\Rightarrow m-1\inƯ\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow m=\left\{2;3;5;0;-1;-3\right\}\)

\(KethopDKXD:m=\left\{2;3;5;0\right\}\)

\(\sqrt[3]{\frac{3}{4}}:\sqrt[3]{\frac{9}{16}}=\frac{3}{4}:\frac{9}{16}\)

\(=\frac{3}{4}.\frac{16}{9}=\frac{4}{3}\)

Học tốt!!!!!!!!!!

ban oi ban lam sai oi

làm hộ mình nha

\(PT\Leftrightarrow x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=1\)

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

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

\(\Leftrightarrow x+\frac{1}{4}=x^4+9x^2+\frac{1}{4}-6x^3-3x+x^2\)

\(\Leftrightarrow x^4-6x^3+10x^2-4x=0\)

\(\Leftrightarrow x\left(x^3-6x^2+10x-4\right)=0\)

Vì x³-6x²+10x-4 \(\ne\)0

nên x=0

Vậy..................................

a) \(\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)

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

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

\(=2\sqrt{2}\)

b) \(\left(1+2\sqrt{3}-\sqrt{2}\right)\left(1+2\sqrt{3}+\sqrt{2}\right)\)

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

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

\(=9+4\sqrt{3}\)