Question

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GIÚP MÌNH NHA _ MÌNH CẢM ƠN^_^

Answer 1

[Lê Thị Thục Hiền]

1)Thay x=4 vào biểu thức B ta được:

\(B=\left(\dfrac{x+1}{2}-\sqrt{x}\right)=\left(\dfrac{4+1}{2}-\sqrt{4}\right)=\dfrac{1}{2}\)

2)\(M=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x+1}{2}-\sqrt{x}\right)\) (đk:\(x\ge0;x\ne1\))

\(=\dfrac{\sqrt{x}+1-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x+1-2\sqrt{x}}{2}\)

\(=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

3) \(M=\dfrac{\sqrt{x}}{6}\) 

=> \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{6}\) \(\Leftrightarrow6\left(\sqrt{x}-1\right)=\sqrt{x}\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow x-5\sqrt{x}+6=0\) \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)  (thỏa)

Vậy...

Answer 2

[LanAnk]

a) \(x=4\rightarrow\sqrt{x}=2\) (TMĐK)

Thay \(\sqrt{x}=2\) vào A ta có :

\(A=\left(\dfrac{1}{2-1}-\dfrac{1}{2+1}\right)=\left(1-\dfrac{1}{3}\right)=\dfrac{2}{3}\)

b) M=A.B

\(\rightarrow M=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right).\left(\dfrac{x+1}{2}-\sqrt{x}\right)\)

\(\rightarrow M=\left(\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(\dfrac{x+1-2\sqrt{x}}{2\sqrt{x}}\right)\)

\(\rightarrow M=\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2\sqrt{x}}\)

\(\rightarrow M=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(M=\dfrac{\sqrt{x}}{6}\)

\(\rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{6}\)

\(\rightarrow6\left(\sqrt{x}-1\right)=\sqrt{x}+1\)

\(\rightarrow6\sqrt{x}-6-\sqrt{x}-1=0\)

\(\rightarrow5\sqrt{x}-7=0\)

\(\rightarrow\sqrt{x}=\dfrac{7}{5}\)

\(\rightarrow x=\pm\dfrac{5\sqrt{7}}{5}\)

\(\rightarrow x=\dfrac{5\sqrt{7}}{7}\) (TMĐK)