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`B=sqrtx/(sqrtx+3)+(2sqrtx)/(\sqrtx-3)-(3x+9)/(x-9)(x>0,x ne 9)`
`=(x-3sqrtx+2x+6sqrtx-3x-9)/(x-9)`
`=(3sqrtx-9)/(x-9)`
`=(3(sqrtx-3))/((sqrtx-3)(sqrtx+3))`
`=3/(sqrtx+3)`
`P=A.B=3/x`
`Px+3\sqrt{x-5}=x-2sqrtx+7(x>=5)`
`<=>3+3\sqrt{x-5}=x-2sqrtx+7`
`<=>x-2sqrtx+4-3\sqrt{x-5}=0`
`<=>2x-4sqrtx+8-6sqrt{x-5}=0`
`<=>x-4sqrtx+4+x-5-6sqrt{x-5}+9=0`
`<=>(sqrtx-2)^2+(\sqrt{x-5}-3)^2=0`
Dấu "=" xảy ra khi $\begin{cases}x=4\\x=14\\\end{cases}(l)$
Vậy khong có giá trị của x thể pt có nghiệm
(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3}{\sqrt{x}+3}.\)
(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{3}\).
Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)
\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)
(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)
\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).
Vậy: \(x\in\varnothing.\)
(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))
\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)
(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)
Vậy: \(x=0.\)
\(a,ĐK:\dfrac{3}{x+7}\ge0\Leftrightarrow x+7>0\left(3>0;x+7\ne0\right)\Leftrightarrow x>-7\\ b,ĐK:\dfrac{-2}{5-x}\ge0\Leftrightarrow5-x< 0\left(2-< 0;5-x\ne0\right)\Leftrightarrow x>5\\ c,ĐK:x^2-7x+10\ge0\Leftrightarrow\left(x-5\right)\left(x-2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le2\end{matrix}\right.\)
\(d,ĐK:x^2-8x+10\ge0\Leftrightarrow\left(x-4-\sqrt{6}\right)\left(x-4+\sqrt{6}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4-\sqrt{6}\ge0\\x-4+\sqrt{6}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4-\sqrt{6}\le0\\x-4+\sqrt{6}\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4+\sqrt{6}\\x\ge4-\sqrt{6}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\)
\(e,ĐK:9x^2+1\ge0\Leftrightarrow x\in R\left(9x^2+1\ge1>0\right)\)
a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)
b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:
\(\left(x-3\right)\left(7-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow3\le x\le7\)
c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:
\(\left(2x-5\right)\left(2x+5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)
d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x>2\)
a: \(P=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\)
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
a:
Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)
b: P>=1/2
=>P-1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)
=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)
=>\(-\sqrt{x}-15>=0\)
=>\(-\sqrt{x}>=15\)
=>căn x<=-15
=>\(x\in\varnothing\)
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>P>=-2
Dấu = xảy ra khi x=0
1) Ta có: \(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}-1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
\(=\dfrac{2x-6\sqrt{x}+x+2\sqrt{x}-3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}-2\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-2}{\sqrt{x}-3}\)
Sửa đề: \(x=7+4\sqrt{3}\)
Thay \(x=7+4\sqrt{3}\) vào P, ta được:
\(P=\dfrac{3\left(2+\sqrt{3}\right)-2}{2+\sqrt{3}-3}=\dfrac{6+3\sqrt{3}-2}{\sqrt{3}-1}\)
\(=\dfrac{4+3\sqrt{3}}{\sqrt{3}-1}=\dfrac{13+7\sqrt{3}}{2}\)
2D : E = | X - 9 | ( ĐKXĐ : X khác 0 )
\(\Leftrightarrow\dfrac{2\left(\sqrt{X}+3\right)}{\sqrt{X}-7}\times\dfrac{\sqrt{X}-7}{2}=\left|X-9\right|\)
\(\Leftrightarrow\sqrt{X}+3=\left|X-9\right|\)
\(\Leftrightarrow\left(\sqrt{X}+3\right)\left|\sqrt{X}-3\right|-\left(\sqrt{X}+3\right)=0\)
\(\Leftrightarrow\left|\sqrt{X}-3\right|-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{X}-3=1\\\sqrt{X}-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{X}=4\\\sqrt{X}=2\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}X=16\\X=4\end{matrix}\right.\)
ĐK : x > 0 và x khác 7 nhé ( sửa )