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:

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

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

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

\(=\sqrt{28y}\)

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

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

\(P=\frac{a\sqrt{a}}{\sqrt{2c+a+b}}+\frac{b\sqrt{b}}{\sqrt{2a+b+c}}+\frac{c\sqrt{c}}{\sqrt{2b+a+c}}\)

\(P=\frac{\sqrt{a^3}}{\sqrt{c+a+b+c}}+\frac{\sqrt{b^3}}{\sqrt{a+a+b+c}}+\frac{\sqrt{c^3}}{\sqrt{b+a+b+c}}\)

\(P=\frac{\sqrt{a^3}}{\sqrt{c+3}}+\frac{\sqrt{b^3}}{\sqrt{a+3}}+\frac{\sqrt{c^3}}{\sqrt{b+3}}\)

dự đoán ra đc a=b=c = 1 với min = 3/2 nhưng chưa biết làm hjhj =))

sửa đề \(B=\frac{a-\sqrt{a}}{a-2\sqrt{a}+1}\left(1-\sqrt{a}\right)\)Với a > 1

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

a, \(A=\left(\frac{x+2}{x-1}-1\right):\left(\frac{-x+2}{x+1}+1\right)\)

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

\(A=\frac{3}{x-1}\cdot\frac{x+1}{3}=\frac{x+1}{x-1}\)

b, x = -5 (thỏa mãn) => \(A=\frac{-5+1}{-5-1}=\frac{-4}{-6}=\frac{2}{3}\)

c, \(A=\frac{1}{4}\Rightarrow\frac{x+1}{x-1}=\frac{1}{4}\)

\(\Rightarrow4x+4=x-1\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=-\frac{5}{3}\) (tm)

\(d,\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\) A thuộc Z <=> 2 chia hết cho x - 1

=> x - 1 thuộc Ư(2)

=> x - 1 thuộc {-1;1;-2;2}

=> x thuộc {0;2;-1;3} mà x khác +-1 

=> x thuộc {0;2;3}