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Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
b: Để A=1/2 thì \(\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10\sqrt{x}+2=\sqrt{x}+3\)
hay \(x\in\varnothing\)
a: \(A=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b: \(B=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1}{x+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\cdot\dfrac{1}{x+\sqrt{x}+1}=\dfrac{-\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(a.R=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}\right)\)
\(R=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+xy-1}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]:\left[\dfrac{xy-1-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]\)
\(R=\dfrac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}+xy-1}{xy-1}:\dfrac{xy-1-x\sqrt{y}+\sqrt{x}+\sqrt{xy}+1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}}{xy-1}\)
\(R=\dfrac{-2\sqrt{x}-2}{xy-1}:\dfrac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)
\(R=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}.\dfrac{xy-1}{-2\left(x\sqrt{y}+\sqrt{xy}\right)}\)
\(R=\dfrac{\sqrt{x}+1}{x\sqrt{y}+\sqrt{xy}}\)
\(b.C=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{7\sqrt{x}+4}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
\(C=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{7\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{2x-6\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(c.M=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+x}=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{\sqrt{x}+x}{\sqrt{x}}=\dfrac{\sqrt{x}+1+x}{\sqrt{x}}\)
\(Q=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-3x}{x-4}\)
\(=\dfrac{-\sqrt{x}}{x-4}\)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
a: \(=\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
c: \(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{x-1}=\dfrac{-x+\sqrt{x}+2}{x-1}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x-1}=\dfrac{-\sqrt{x}+2}{\sqrt{x}-1}\)
\(M=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)
\(\Leftrightarrow M=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(\Leftrightarrow M=x-\sqrt{x}-x-\sqrt{x}+x+1=x-2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)