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\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
\(A=\left(\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}\right)\left(\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}\right)\)\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)=a-1\)
Do \(A=-a^2\Rightarrow a-1=-a^2\)=> \(a^2+a-1=0=>4a^2+4a+1-5=0=>\left(2a+1\right)^2=5\) Xét 2a+1=-5 và 5 là ra
a) \(A=\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right)=\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)=a-1\)
b) Ta có : \(a-1=-a^2\Leftrightarrow a^2+a-1=0\) \(\Leftrightarrow\orbr{\begin{cases}a=\frac{-1-\sqrt{5}}{2}\left(\text{loại}\right)\\a=\frac{-1+\sqrt{5}}{2}\left(\text{nhận}\right)\end{cases}}\)
\(A=\)\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}\)\(:\)\(\left[\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\)\(\left(1+a+2\sqrt{a}\right)\left(1+a-2\sqrt{a}\right)\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(1+a\right)\left[\left(1+a\right)^2-\left(2\sqrt{a}\right)^2\right]}\)\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1+2a+a^2-4a\right)}\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1-a\right)^2}=\frac{\sqrt{q}}{a+1}\)
ĐKXĐ:a,b>=0 ;a khác b
=\(\left(\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right).\frac{1}{a-b}\) +\(\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
=\(\left(\sqrt{ab}-\sqrt{ab}\right).\frac{1}{a-b}\)+\(\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
=0..\(\frac{1}{a-b}\)+\(\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
=\(\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
vậy.......................
k mk nha
\(P=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(P=\left(a+2\sqrt{a}+1\right).\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(P=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(P=1\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}-2}{\sqrt{a}+1}\)
\(=\frac{a-2}{\sqrt{a}\left(a-1\right)}\)
tick cho mình nha bạn