cho A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\) - \(\dfrac{5}{x+\sqrt{x}-6}\)+\(\dfrac{1}{2-\sqrt{x}}\)
a,Rút gọn A
b,TÌm x để A có giá trị nguyên
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Những câu hỏi liên quan
18 tháng 8 2021
a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:
\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)
b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)
\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)
19 tháng 8 2021
c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)
hay \(x\in\left\{16;25;64\right\}\)
2 tháng 10 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)
b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)
\(\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
27 tháng 10 2023
a: Khi x=25 thì \(A=\dfrac{5-2}{5-3}=\dfrac{3}{2}\)
b: P=A*B
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(\dfrac{6x+6\sqrt{x}-12}{x+5\sqrt{x}+4}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\left(\dfrac{6x+6\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\dfrac{6x+6\sqrt{x}-12-5x-5\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
c: \(\sqrt{P}< =\dfrac{1}{2}\)
=>0<=P<=1/4
=>\(\left\{{}\begin{matrix}P>=0\\P-\dfrac{1}{4}< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{1}{4}< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{4\left(\sqrt{x}-2\right)-\sqrt{x}+1}{4\left(\sqrt{x}-1\right)}< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{3\sqrt{x}-7}{\sqrt{x}-1}< =0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< \sqrt{x}< =\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< x< \dfrac{49}{9}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\x=\dfrac{49}{9}\end{matrix}\right.\)
=>\(4< =x< =\dfrac{49}{9}\)
mà x nguyên
nên \(x\in\left\{4;5\right\}\)
16 tháng 11 2021
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
18 tháng 10 2021
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
Đkxđ : \(x\ne4\); \(x\ge0\)
a.\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}\right)^2+3\sqrt{x}-4\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)-4\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
b. Ta có : \(A=\) \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
\(\Leftrightarrow A\left(\sqrt{x}-2\right)=\sqrt{x}-4\)
\(\Leftrightarrow A\sqrt{x}-2A=\sqrt{x}-4\)
\(\Leftrightarrow A\sqrt{x}-\sqrt{x}=2A-4\)
\(\Leftrightarrow\sqrt{x}\left(A-1\right)=2\left(A-2\right)\)
\(\Leftrightarrow\sqrt{x}=\dfrac{2\left(A-2\right)}{A-1}\)
\(\Leftrightarrow x=\left(\dfrac{2\left(A-2\right)}{A-1}\right)^2\)
Vậy .....
Chúc bạn học tốt ạ !
: ĐKXĐ : x≥0x≥0 và x≠4x≠4
Câu a : A=√x+2√x+3−5x+√x−6+12−√xA=x+2x+3−5x+x−6+12−x
=(√x+2)(√x−2)(√x−2)(√x+3)−5(√x−2)(√x+3)−√x+3(√x−2)(√x+3)=(x+2)(x−2)(x−2)(x+3)−5(x−2)(x+3)−x+3(x−2)(x+3)
=(√x+2)(√x−2)−5−(√x+3)(√x−2)(√x+3)=(x+2)(x−2)−5−(x+3)(x−2)(x+3)
=x−4−5−√x+3(√x−2)(√x+3)=x−4−5−x+3(x−2)(x+3)
=x−√x−6(√x−2)(√x+3)=x−x−6(x−2)(x+3)
=(√x+2)(√x−3)(√x−2)(√x+3)=(x+2)(x−3)(x−2)(x+3)
Câu c :
A<1A<1 ⇔(√x+2)(√x−3)(√x−2)(√x+3)<1⇔(x+2)(x−3)(x−2)(x+3)<1
⇔(√x+2)(√x−3)<(√x−2)(√x+3)⇔(x+2)(x−3)<(x−2)(x+3)
⇔x−√x−6<x+√x−6⇔x−x−6<x+x−6
⇔−2√x<0⇔−2x<0 ( Luôn đúng với mọi x khi {x>0x≠4{x>0x≠4)
Vậy các giá của x để A < 1 là {x>0x≠4