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

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

\(=\frac{\sqrt{5}-1}{5-1}+\frac{\sqrt{5}+2}{5-4}-\frac{3+\sqrt{5}}{9-5}-\sqrt{5}\)

\(=\frac{\sqrt{5}-1}{4}+\sqrt{5}+2-\frac{3+\sqrt{5}}{4}-\sqrt{5}\)

\(=\frac{1}{2}+2=\frac{5}{2}\)

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

\(=\frac{\sqrt{5}-1}{4}+\sqrt{5}+2-\frac{3+\sqrt{5}}{4}-\sqrt{5}\)

\(=\frac{\sqrt{5}-1-3-\sqrt{5}}{4}+2=-1+2=1\)

b14:

\(a,P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)

\(P=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)

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

sao ko gọn zị :v

\(M=\frac{3\left(\sqrt{x}+3\right)-8}{\sqrt{x}+3}=3-\frac{8}{\sqrt{x}+3}\)

Để M nguyên thì \(\frac{8}{\sqrt{x}+3}\)nguyên hay \(\sqrt{x}+3\inƯ\left(8\right)\)

bạn lập bảng xét nhé ;)

Bài 10 : 

a, Thay x = 16 vào P ta được : \(P=\frac{2\sqrt{16}-3}{\sqrt{16}+2}=\frac{2.4-3}{4+2}=\frac{5}{6}\)

b, \(P=\frac{2\sqrt{x}-3}{\sqrt{x}+2}=\frac{2\left(\sqrt{x}+2\right)-7}{\sqrt{x}+2}=2-\frac{7}{\sqrt{x}+2}\)

\(\Rightarrow\sqrt{x}+2\inƯ\left(7\right)=\left\{1;7\right\}\)

\(M=\frac{\sqrt{x}+1+4}{\sqrt{x}+1}=1+\frac{4}{\sqrt{x}+1}\)

Để M nguyên thì \(\frac{4}{\sqrt{x}+1}\)nguyên hay \(\sqrt{x}+1\inƯ\left(4\right)\)( tự lập bảng xét )

\(P=\frac{\sqrt{x}+2+5}{\sqrt{x}+2}=1+\frac{5}{\sqrt{x}+2}\)tương tự ý trên

10 Tú làm rồi không làm lại

Bài 4:

Để M nguyên \(\Leftrightarrow\sqrt{x}+2⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}+4⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3+7⋮2\sqrt{x}-3\)

Mà \(\Leftrightarrow2\sqrt{x}-3⋮2\sqrt{x}-3\)

\(\Rightarrow7⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Mà \(2\sqrt{x}-3\ge-3\left(x\ge0\right)\)

\(\Rightarrow2\sqrt{x}-3\in\left\{1;-1;7\right\}\)

\(\Rightarrow x\in\left\{4;1;25\right\}\)

Vậy...

các phần khác tương tự nhé

a, Với \(x\ge0;x\ne9\)

\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right)\)

\(=\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right)\)

\(=\frac{-3\sqrt{x}-3}{x-9}\)vậy ko xảy ra đpcm

bạn giải thích rõ cho mk đc ko

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}-\frac{36+12\sqrt{6}}{3}\right)\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)

a, Với \(x\ge0;x\ne4;9\)

\(Q=\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}+\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}\)

\(=\frac{\sqrt{x}+2+x-9-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{1}{\sqrt{x}-2}\)

b,\(A=\frac{P}{Q}\Rightarrow\frac{1}{\sqrt{x}+1}.\left(\sqrt{x}-2\right)=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

\(\Rightarrow A< 0\)vì \(\left|A\right|\ge0\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< 0\Rightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)

Kết hợp với đk vậy  \(0\le x< 4\)mà x phải là số nguyên tố => x = 1 ; x = 3 

đúng nha bạn 

Trả lời:

\(2y\sqrt{\frac{7}{y}}\left(y>0\right)\)

\(=\sqrt{\left(2y\right)^2.\frac{7}{y}}\)

\(=\sqrt{4y^2.\frac{7}{y}}\)

\(=\sqrt{28y}\)

\(\sqrt{48y^8}=\sqrt{48}.\sqrt{y^8}=4\sqrt{3}y^4\)

\(\sqrt{48y^8}=\sqrt{16\cdot3\cdot y^8}=\left|16\right|\cdot\left|y^4\right|\cdot\sqrt{3}=4\sqrt{3}y^4\)