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Bài 3:
\(C=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1+2}{a-1}\)
\(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{a-1}{\sqrt{a}+3}\)
\(=\dfrac{\left(a-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
B=\(\dfrac{\sqrt{a.6}}{\sqrt{6.6}}+\dfrac{\sqrt{2a.3}}{\sqrt{3.3}}+\dfrac{\sqrt{3a.2}}{\sqrt{2.2}}\)
=\(\dfrac{\sqrt{6a}}{6}+\dfrac{\sqrt{6a}}{3}+\dfrac{\sqrt{6a}}{2}\)
=\(\dfrac{\sqrt{6a}+2\sqrt{6a}+3\sqrt{6a}}{6}\)
=\(\dfrac{6\sqrt{6a}}{6}=\sqrt{6a}\)
b: \(B=\dfrac{\sqrt{6}}{6}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{3}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{2}\cdot\sqrt{a}\)
\(=\sqrt{a}\cdot\sqrt{6}=\sqrt{6a}\)
e: \(=2-x-x=2-2x\)
i: \(=\left|x-\left(1-x\right)\right|-2x=\left|x-1+x\right|-2x\)
\(=\left|2x-1\right|-2x\)
=1-2x-2x=1-4x
Bài 1:
A.\(\left(\sqrt{x}+2\right)\) = -1 (ĐK: \(x\ge0\)
\(\Leftrightarrow\dfrac{1}{x-4}\left(\sqrt{x}+2\right)=-1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-2}=-1\)
\(\Leftrightarrow\sqrt{x}-2=-1\)
\(\Leftrightarrow\sqrt{x}=1\\ \Leftrightarrow x=1\left(TM\right)\)
Vậy x = 1
Bài 2: ĐK: \(x\ge0\)
Để \(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3\right\}\)\(\Leftrightarrow x\in\left\{1\right\}\)
Bài 3:
a, Ta có: \(x+\sqrt{x}+1=x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}+1\\ =\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Ta có: 2 > 0 và \(x+\sqrt{x}+1>0\Rightarrow C>0\) và \(x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(C=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có: \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow x=0\left(TM\right)\)
Vậy MaxC = 2 khi x = 0
Còn cái GTNN chưa tính ra được, để sau nha
Bài 4: ĐK: \(x\ge0,x\ne1\)
\(D=\left(\dfrac{2x+1}{\sqrt{x^3-1}}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\sqrt{x}-1\)
\(D=3\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=2\left(TM\right)\)
\(D=x-3\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}-1=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(1-\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=9\left(TM\right)\end{matrix}\right.\)
Bài 5: \(E< -1\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}+1< 0\Leftrightarrow\dfrac{-3x+2x+4\sqrt{x}}{2x+4\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-x}{2x+4\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)
Ta có: \(\sqrt{x}>0\Leftrightarrow x>0\Leftrightarrow2x+4\sqrt{x}>0\) mà \(\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)\(\Rightarrow\sqrt{x}\left(4-\sqrt{x}\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\left(L\right)\\4-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>0\\4-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\0< x< 16\end{matrix}\right.\)
a: \(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
b: Để B=7 thì \(x+\sqrt{x}+1=\dfrac{2}{7}\)
\(\Leftrightarrow x+\sqrt{x}+\dfrac{5}{7}=0\)
hay \(x\in\varnothing\)
Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)
\(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-\left(x+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+1-x-2-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+0-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left[-\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(-1\right)}{x+\sqrt{x}+1}\\ =-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
Bài 3:
\(P=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\left(2x+\sqrt{x}\right)\sqrt{x}}{x}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+2\left(\sqrt{x}+1\right)\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x\left(2\sqrt{x}+1\right)}{x}+2\sqrt{x}+2\)
\(=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+1\\ =\dfrac{x-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{2x+1}{x+\sqrt{x}+1}\)
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
a: \(A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\)
\(=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+2\sqrt{x}}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{x\sqrt{x}+x+\sqrt{x}+1+x-\sqrt{x}+2\sqrt{x}}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{x\sqrt{x}+2x+2\sqrt{x}+1}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)+2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b: Để A<0 thì \(\sqrt{x}-1< 0\)
=>0<x<1