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TL:
\(A=\frac{\sqrt{x+2}}{\sqrt{x-5}}\) mà x = 9
\(A=\frac{\sqrt{0+2}}{\sqrt{9-2}}\)
\(A=\frac{\sqrt{11}}{2}\)
b) chưa bt làm
a/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
Thay \(x=9\) vào biểu thức ta có :
\(A=\frac{\sqrt{9}+2}{\sqrt{9}-5}=\frac{3+2}{3-5}=-\frac{5}{2}\)
Vậy....
b/ Ta có :
\(B=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{x-25}\)
\(=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{3\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-5\right)}+\frac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{1}{\sqrt{x}-5}\)
Vậy...
c/ Ta có :
\(A=B.\left|x-4\right|\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{1}{\sqrt{x}-5}\left|x-4\right|\)
\(\Leftrightarrow\sqrt{x}+2=\left|x-4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=x-4\\\sqrt{x}+2=4-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}-6=0\\x+\sqrt{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\\\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Vậy...
Câu a bạn tự giải
\(B=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{3\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\frac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{1}{\sqrt{x}-5}\)
a/ khi x = 9 thì A = \(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{5}{-2}=-\dfrac{5}{2}\)
b/ B = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{1}{\sqrt{x}-5}\left(đpcm\right)\)
c/ \(A=B\cdot\left|x-4\right|\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-5}=\dfrac{1}{\sqrt{x}-5}\cdot\left|x-4\right|\)
\(\Leftrightarrow\left|x-4\right|=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)
Vì: \(\sqrt{x}+2>0\)=> đk: x > 4
\(\left|x-4\right|=\sqrt{x}+2\)
\(\Leftrightarrow x-4=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-6=0\)
\(\Leftrightarrow\left(x-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{25}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\dfrac{1}{2}=\dfrac{5}{2}\\\sqrt{x}-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-2\left(loai\right)\end{matrix}\right.\)
\(\sqrt{x}=3\Leftrightarrow x=9\left(TM\right)\)
Vậy x = 9 thì A = B.|x - 4|
a: \(M=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Khi a=9/25 thì \(M=\dfrac{\dfrac{3}{5}-4}{\dfrac{3}{5}-2}=\dfrac{-17}{5}:\dfrac{-7}{5}=\dfrac{17}{7}\)
c: Để |M|=1/6 thì M=1/6 hoặc M=-1/6
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=\dfrac{1}{6}\\\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6\sqrt{a}-24=\sqrt{a}-2\\6\sqrt{a}-24=-\sqrt{a}+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5\sqrt{a}=22\\7\sqrt{a}=26\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\left(\dfrac{22}{5}\right)^2\\a=\left(\dfrac{26}{7}\right)^2\end{matrix}\right.\)
học lớp 9 chưa mà đòi đăng ? :))
a) Ta có : \(A=\frac{x+5\sqrt{x}}{x-25}=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)
Để A nhận giá trị = 0 thì \(\sqrt{x}=0\)<=> x = 0 ( tmđk )
Vậy với x = 0 thì A = 0
b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
c) P = B : A = \(\frac{\frac{\sqrt{x}}{\sqrt{x}+3}}{\frac{\sqrt{x}}{\sqrt{x}-5}}=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{\sqrt{x}}{\sqrt{x}+3}\times\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
Xét hiệu P - 1 ta có :
\(\frac{\sqrt{x}-5}{\sqrt{x}+3}-1=\frac{\sqrt{x}-5}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\frac{-8}{\sqrt{x}+3}\)
Vì \(\hept{\begin{cases}-8< 0\\\sqrt{x}+3>0\end{cases}}\Rightarrow\frac{-8}{\sqrt{x}+3}< 0\)hay P - 1 < 0
=> P < 1
a) \(A=0\Rightarrow\frac{x+5\sqrt{x}}{x-25}=0\Rightarrow x+5\sqrt{x}=0\Leftrightarrow x=0\)(thỏa mãn).
b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)
\(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}\)
c) \(P=B\div A=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}=1-\frac{8}{\sqrt{x}+3}< 1\)
Với x ≥ 0 , x ≠ 25 thì B = 3 x + 5 + 20 − 2 x x − 15 = 3 x + 5 + 20 − 2 x x + 5 x − 5
= 3 x − 5 + 20 − 2 x x + 5 x − 5 = 3 x − 15 + 20 − 2 x x + 5 x − 5 = x + 5 x + 5 x − 5 = 1 x − 5
(điều phải chứng minh)