Cho A = \(\frac{x+3}{-2}\)và B = \(\frac{-9}{x+3}\). Tìm x để A = 2B
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15 tháng 12 2017
a, ĐKXĐ : x^2-9 khác 0 ; x-3 khác 0 ; x+3 khác 0 => x khác -3 và 3
A = x^2+3+2.(x-3)-(x+3)/(x-3).(x+3) = x^2+x-6/(x-3).(x+3) = (x-2).(x+3)/(x-3).(x+3) = x-2/x-3
b, Để A = 1/2 => x-2 = 2.(x-3) = 2x-6
=> x = 4 (tm ĐKXĐ)
k mk nha
AH
Akai Haruma
Giáo viên
9 tháng 7 2020
Lời giải:
a) ĐK: $x>0; x\neq 9$
Ta có:
\(A=\frac{\sqrt{x}+15}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{x}{\sqrt{x}(\sqrt{x}-3)}+\frac{2\sqrt{x}+5}{\sqrt{x}+3}\)
\(A=\frac{\sqrt{x}+15}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\frac{(2\sqrt{x}+5)(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\)
\(=\frac{\sqrt{x}+15-x-3\sqrt{x}+2x-\sqrt{x}-15}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{x-3\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
b)
\(A=2B\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}+3}=\frac{2(\sqrt{x}-3)}{14}=\frac{\sqrt{x}-3}{7}\)
\(\Rightarrow 7\sqrt{x}=x-9\Leftrightarrow x-7\sqrt{x}-9=0\)
\(\Rightarrow \sqrt{x}=\frac{7+\sqrt{85}}{2}\Leftrightarrow x=\frac{67+7\sqrt{85}}{2}\)
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
KN
29 tháng 7 2019
a) \(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x^2-6x}{\left(x+3\right)\left(x-3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{3x^2-13x}{x^2-9}\)
14 tháng 10 2020
\(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
a) ĐK : x ≠ ±3
\(=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)
b) Để A < 2
=> \(\frac{3x}{x-3}< 2\)
<=> \(\frac{3x}{x-3}-2< 0\)
<=> \(\frac{3x}{x-3}-\frac{2x-6}{x-3}< 0\)
<=> \(\frac{3x-2x+6}{x-3}< 0\)
<=> \(\frac{x+6}{x-3}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}x+6>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-6\\x< 3\end{cases}}\Leftrightarrow-6< x< 3\)
2. \(\hept{\begin{cases}x+6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -6\\x>3\end{cases}}\)( loại )
Vậy -6 < x < 3
3 tháng 4 2021
a, \(A=\left(\frac{x}{x+3}+\frac{x}{x-3}-\frac{2}{x^2-9}\right).\frac{x+3}{2x-2}\)
\(=\frac{x^2-3x+x^2+3x-2}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{2\left(x-1\right)}=\frac{2\left(x-1\right)\left(x+1\right)\left(x+3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+3\right)}=\frac{x+1}{x-3}\)
Ta có : A = 2 hay \(\frac{x+1}{x-3}=2\Rightarrow x+1=2x-6\Leftrightarrow-x=-7\Leftrightarrow x=7\)(tmđk )
b, \(A< 0\Rightarrow\frac{x+1}{x-3}< 0\)
TH1 : \(\hept{\begin{cases}x+1< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>3\end{cases}}}\)( vô lí )
TH2 : \(\hept{\begin{cases}x+1>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow-1< x< 3}}\)
Kết hợp với đk ta được -1 < x < 3 ; x khác 1
Để A = 2B nên:
\(\frac{x+3}{-2}=\frac{-18}{2\left(x+3\right)}\)
\(\Rightarrow\left(x+3\right).2\left(x+3\right)=-18.\left(-2\right)\)
\(\Rightarrow2.\left(x+3\right)^2=36\)
\(\Rightarrow2x^2+2.3^2=36\)
\(\Rightarrow2x^2+18=36\)
\(\Rightarrow2x^2=18\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x=3\)
Vậy A = 2B khi x = 3
Để A = 2B nên :
\(\frac{x+3}{-2}=\frac{-18}{2\left(x+3\right)}\)
\(\Rightarrow\left(x+3\right).2\left(x+3\right)=-18.\left(-2\right)\)
\(\Rightarrow2.\left(x+3\right)^2=36\)
\(\Rightarrow2x^2-2.3^2=36\)
\(\Rightarrow2x^2+18=36\)
\(\Rightarrow2x^2=18\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x=3\)
Vậy x = 3