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

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

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

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}

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

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

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

\(B=\left|\sqrt{a-1}+1\right|+\left|\sqrt{a-1}-1\right|\)

có \(a\le2\Rightarrow a-1\le1\Rightarrow\sqrt{a-1}\le1\Rightarrow\sqrt{a-1}-1\le0\)

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

Đây nè bạn!!

\(A=\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}\)

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

\(A=\left|a+3\right|+\left|a-3\right|\)

có \(\hept{\begin{cases}a\ge-3\Rightarrow a+3\ge0\\a\le3\Rightarrow a-3̸\le0\end{cases}}\)

nên \(A=a+3+3-a=6\)

a, Để A có nghĩa \(x^2-1\ge0\Leftrightarrow\left(x-1\right)\left(x+1\right)\ge0\Leftrightarrow x\le-1;x\ge1\)

b,  \(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)với \(x\ge\sqrt{2}\)

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

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

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