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Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A = \(\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(\Leftrightarrow A=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1}{2\left(x-1\right)}-\dfrac{\sqrt{x}-1}{2\left(x-1\right)}-\dfrac{2\sqrt{x}}{2\left(x-1\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(x-1\right)}\)
\(\Leftrightarrow A=\dfrac{2\left(1-\sqrt{x}\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{1}{\sqrt{x}+1}\)
b) Khi \(x=\dfrac{4}{9}\) (thảo mãn ĐKXĐ) thì giá trị của A là:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{\dfrac{4}{9}}+1}=-\dfrac{3}{5}\)
Vậy .....
c)
+) Khi \(A=-\dfrac{1}{2}\) thì ta có:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=1\) (Loại do không thỏa mãn ĐKXĐ)
+) Khi \(A=\dfrac{-1}{4}\) thì ta có:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{4}\)
\(\Leftrightarrow x=9\) (thỏa mãn)
Vậy để A = \(-\dfrac{1}{4}\) thì x = 9
a/ ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
= \(\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}+\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{2-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{-1}{\sqrt{x}+1}\)
b/
Thay \(x=\dfrac{4}{9}\) vào A ta được:
\(A=\dfrac{-1}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-1}{\dfrac{2}{3}+1}=\dfrac{-3}{5}\)
Vậy khi \(x=\dfrac{4}{9}\) thì \(A=\dfrac{-3}{5}\)
c/ Với \(x\ge0,x\ne1\)
* Để \(A=\dfrac{-1}{2}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{2}\)
\(\Leftrightarrow-2=-\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\) ( ktmđk)-Loại
Vậy không có giá trị nào của x thỏa mãn \(A=\dfrac{-1}{2}\)
* Để \(A=\dfrac{-1}{4}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{4}\)
\(\Leftrightarrow-4=-\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\) (tmđk)
Vậy để \(A=\dfrac{-1}{4}\) thì \(x=9\)
\(2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}=\left[{}\begin{matrix}2x-1-1=2x-2khix-5>0\\2x-1+1=2xkhix-5< 0\end{matrix}\right.\)
b) \(\dfrac{\sqrt{x^2-4x+4}}{x^2-2}=\dfrac{\sqrt{\left(x-2\right)^2}}{x^2-2}=\left[{}\begin{matrix}\dfrac{x-2}{x^2-2}khix-2\ge0\\\dfrac{2-x}{x^2-2}khix-2\le0\end{matrix}\right.\)
\(A=\dfrac{x^2-x^2-2x+2x-4}{\left(x-2\right)\left(x+2\right)}\left(x\ne\pm2\right)=\dfrac{-4}{x^2-4}\\ B=\dfrac{\left|x-1\right|+x^2+4x+4}{\left(x-2\right)\left(x+2\right)}\left(x\ne\pm2\right)\)
Với \(x>1;x\ne2\Leftrightarrow B=\dfrac{x^2+5x+3}{\left(x-2\right)\left(x+2\right)}\)
Với \(x< 1;x\ne-2\Leftrightarrow B=\dfrac{x^2+3x+5}{\left(x-2\right)\left(x+2\right)}\)