Tìm `x` để `AB. B > 3/2`, biết:
\(A=\dfrac{3+\sqrt{x}}{\sqrt{x}}\)
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
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Những câu hỏi liên quan
29 tháng 10 2021
\(C=\left(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}+4}\)
\(=\dfrac{-3}{2\sqrt{x}+4}\)
Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)
\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)
\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)
hay \(0\le x< \dfrac{25}{4}\)
10 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(B=\dfrac{x-3}{x-1}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x-3-2\left(\sqrt{x}-1\right)+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}-2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(ĐặtP=\dfrac{A}{B}\)
=>\(P=\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}-2}{\sqrt{x}}\)
Để P<1 thì P-1<0
=>\(\dfrac{2\sqrt{x}-2-\sqrt{x}}{\sqrt{x}}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên
nên \(x\in\left\{0;1;2;3\right\}\)
28 tháng 10 2021
a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3}{\sqrt{x}-3}\)
T
30 tháng 8 2023
Ta có: \(P=A\cdot B\) (ĐK: \(x>0;x\ne4\))
\(=\left(\dfrac{3\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}-\dfrac{1}{2-\sqrt{x}}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left[\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right]\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left(\dfrac{3+\sqrt{x}-3}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left(1+\dfrac{1}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+9}\)
Với x > 0; x ≠ 4 thì \(\sqrt{P}< \dfrac{1}{3}\Leftrightarrow P< \dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}< \dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}-\dfrac{1}{9}< 0\)
\(\Leftrightarrow\dfrac{9\left(\sqrt{x}-1\right)}{9\left(\sqrt{x}+9\right)}-\dfrac{\sqrt{x}+9}{9\left(\sqrt{x}+9\right)}< 0\)
\(\Leftrightarrow\dfrac{9\sqrt{x}-9-\sqrt{x}-9}{9\sqrt{x}+81}< 0\)
\(\Leftrightarrow\dfrac{8\sqrt{x}-18}{9\sqrt{x}+18}< 0\)
Ta thấy: \(9\sqrt{x}+18>0\forall x\)
\(\Rightarrow8\sqrt{x}-18< 0\)
\(\Rightarrow\sqrt{x}< \dfrac{18}{8}\)
\(\Rightarrow\sqrt{x}< \dfrac{9}{4}\Leftrightarrow x< \dfrac{81}{16}\)
Kết hợp với điều kiện, ta được: \(0< x\le5\)\(;x\ne4\)
\(\Rightarrow x\in\left\{1;2;3;5\right\};x\in Z\) thì \(\sqrt{P}< \dfrac{1}{3}\)
#Urushi
26 tháng 7 2021
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)
=2
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
31 tháng 10 2021
\(a,=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}+3}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}\)
31 tháng 10 2021
a: \(=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
15 tháng 1 2022
a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)
b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)
=>x<1
17 tháng 10 2021
\(a,A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
Tìm \(x\) để \(A.B>\dfrac{3}{2}?\)
\(ĐK:x>0;\\ A.B\\ =\dfrac{3+\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =1+\dfrac{1}{\sqrt{x}}\\ A.B>\dfrac{3}{2}\\ \Leftrightarrow1+\dfrac{1}{\sqrt{x}}>\dfrac{3}{2}\\ \Leftrightarrow\dfrac{1}{\sqrt{x}}>\dfrac{3}{2}-1\\ \Leftrightarrow\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\\ \Leftrightarrow\sqrt{x}< 2\\ \Leftrightarrow0< x< 4\)
Vậy \(0< x< 4\) thì \(A.B>\dfrac{3}{2}\)