(3) cho bt P= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)
a) rút gọn P
b) tìm điều kiện của x để P >0
c) tìm x nguyên để P nhận giá trị nguyên
giúp mk vs ạ
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HP
31 tháng 8 2021
a, ĐK: \(x\ge0;x\ne9\)
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=-\dfrac{3}{\sqrt{x}-3}\)
HP
31 tháng 8 2021
b, \(P>0\Leftrightarrow-\dfrac{3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3>0\)
\(\Leftrightarrow x>9\)
c, \(P=-\dfrac{3}{\sqrt{x}-3}\in Z\)
\(\Leftrightarrow\sqrt{x}-3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)
\(\Leftrightarrow x\in\left\{0;4;16;36\right\}\)
16 tháng 10 2021
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
AM
26 tháng 12 2021
\(A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{2x-6\sqrt{x}+x+\sqrt{x+}3\sqrt{x}+3+3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{3x-13\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
19 tháng 10 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}-\dfrac{4\sqrt{x}}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
31 tháng 8 2023
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
NB
23 tháng 6 2021
a) \(P=\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)
\(\left(x\ge0;x\ne4;9\right)\)
b)\(P=-1\Leftrightarrow4x+\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
c) bpt đưa về dạng \(4mx>x+1\Leftrightarrow\left(4x-1\right)x>1\)
Nếu \(4m-1\le0\) thì tập nghiệm không thể chứa mọi giá trị \(x>9\); Nếu \(4m-1>0\) thì tập nghiệm bpt là \(x>\dfrac{1}{4m-1}\). Do đó bpt tm mọi \(x>9\Leftrightarrow9\ge\dfrac{1}{4m-1}\) và \(4m-1>0\). ta có \(m\ge\dfrac{5}{18}\)
24 tháng 8 2021
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
VT
19 tháng 10 2021
a) Tại x=16 thì A = \(\dfrac{\sqrt{16}-1}{\sqrt{16}+2}=\dfrac{4-1}{4+2}=\dfrac{1}{2}\)
b) B = \(\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\div\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
= \(\dfrac{\sqrt{x}+1+x-\sqrt{x}}{x+\sqrt{x}}\times\dfrac{x+\sqrt{x}}{\sqrt{x}}\)
= \(\dfrac{x+1}{\sqrt{x}}\)
B = \(\dfrac{x+1}{\sqrt{x}}\)= 2
⇒ x + 1 = 2\(\sqrt{x}\)
⇒ x - \(2\sqrt{x}\) +1 = 0
⇒ \(\left(\sqrt{x}-1\right)^2\) = 0
⇒ \(\sqrt{x}-1=0\)
⇒ x = 1
a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3}{\sqrt{x}-3}\)