hepl
a: \(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)
\(=\frac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)
Ta có: \(B=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b: \(x=\sqrt{\sqrt5-\sqrt{3-\sqrt{29-12\sqrt5}}}\)
\(=\sqrt{\sqrt5-\sqrt{3-\sqrt{20-2\cdot2\sqrt5\cdot3+9}}}\)
\(=\sqrt{\sqrt5-\sqrt{3-\sqrt{\left(2\sqrt5-3\right)^2}}}\)
\(=\sqrt{\sqrt5-\sqrt{3-\left(2\sqrt5-3\right)}}=\sqrt{\sqrt5-\sqrt{6-2\sqrt5}}\)
\(=\sqrt{\sqrt5-\sqrt{\left(\sqrt5-1\right)^2}}\)
\(=\sqrt{\sqrt5-\left(\sqrt5-1\right)}=\sqrt1=1\)
Khi x=1 thì B không có giá trị
c: B>0
=>\(-\sqrt{x}\left(\sqrt{x}-1\right)>0\)
=>\(-\sqrt{x}+1>0\)
=>\(-\sqrt{x}>-1\)
=>\(\sqrt{x}<1\)
=>0<=x<1
d: \(B=-x+\sqrt{x}\)
\(=-\left(x-\sqrt{x}\right)\)
\(=-\left(x-\sqrt{x}+\frac14-\frac14\right)=-\left(\sqrt{x}-\frac12\right)^2+\frac14\le\frac14\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}-\frac12=0\)
=>\(\sqrt{x}=\frac12\)
=>\(x=\frac14\) (nhận)
