Bài 1:
a: ĐKXĐ: x>=0
\(M=x+\sqrt{x}-1\)
\(=x+\sqrt{x}+\frac14-\frac54\)
\(=\left(\sqrt{x}+\frac12\right)^2-\frac54\ge\frac14-\frac54=-1\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
b: \(N=-x+\sqrt{x-1}+2\)
=\(-x+1+\sqrt{x-1}+1\)
\(=-\left(\sqrt{x-1}\right)^2+\sqrt{x-1}-\frac14+\frac54=-\left(\sqrt{x-1}-\frac12\right)^2+\frac54\le\frac54\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x-1}-\frac12=0\)
=>\(x-1=\frac14\)
=>\(x=1+\frac14=\frac54\) (nhận)
Bài 2:
a: \(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\cdot\sqrt{x}-\sqrt{x}+x-1}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{2}{\left(\sqrt{x}+1\right)^2}=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)
\(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\)
\(=\frac{1}{\sqrt{x}-1}-\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{1}{\sqrt{x}+1}\)
Ta có: \(Q=\left(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\cdot\sqrt{x}-\sqrt{x}+x-1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\right)\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\cdot\left(\sqrt{x}+1\right)=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}+1\ge1\forall x\) thỏa mãn ĐKXĐ
=>\(\frac{2}{\sqrt{x}+1}\le\frac21=2\forall x\) thỏa mãn ĐKXĐ
=>\(-\frac{2}{\sqrt{x}+1}\ge-2\forall x\) thỏa mãn ĐKXĐ
=>\(-\frac{2}{\sqrt{x}+1}+1\ge-2+1=-1\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
Bài 3:
a: \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: \(P=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(=\frac{-5\sqrt{x}-15+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\le\frac{17}{3}-5=\frac23\forall x\)
Dấu '=' xảy ra khi x=0
