\(\text{Mình cũng muốn xem đáp àn bài này là bao nhêu}\)

\(\text{Chững nào có đáp án Nguyễn Minh Hiển nhắn tin với mình nha}\)

Vì M1 là trung điểm của MA, mà M là trung điểm của AB => \(M1A=\frac{1}{4}AB\)

M1000A = \(1\times\left(\frac{1}{4}\div\left(2\times1000\right)\right)=0,000125\left(m\right)\)

\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{5}{x+\sqrt{x}-6}-\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{\sqrt{x}+3}{x+\sqrt{x}-6}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{5}-1\Rightarrow P=\frac{\sqrt{5}-5}{\sqrt{5}-3}\)

\(P=\frac{2}{3}\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-2}=\frac{2}{3}\Leftrightarrow3\sqrt{x}-12=2\sqrt{x}-4\Leftrightarrow\sqrt{x}=8\Leftrightarrow x=64\)

\(d,\text{chua nghi ra :(}\)

\(e;P>1\)

\(+,\sqrt{x}>2\Leftrightarrow x>4\Leftrightarrow\sqrt{x}-2>0ma:\left\{{}\begin{matrix}\sqrt{x}-2>\sqrt{x}-4\\\sqrt{x}-2>0\end{matrix}\right.\Rightarrow P< 1\left(loai\right)\)

\(+,\sqrt{x}< 2\Leftrightarrow x< 4\Leftrightarrow\sqrt{x}-2< 0ma:\left\{{}\begin{matrix}\sqrt{x}-4< \sqrt{x}-2\\\sqrt{x}-2< 0\end{matrix}\right.\Rightarrow P>1\left(\text{thoaman}\right)\)

(126+32) x (18-16-2)

= 158      x       0

= 0

\(a;Đk:x\ne1\)

\(A=\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2}{x-1}=\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x}+1}{x-1}-\frac{2}{x-1}=\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{\sqrt{x}+1}\) \(b;x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\Rightarrow A=\frac{2}{\sqrt{2}}=\sqrt{2}\)

\(c;A=\frac{1}{4}=\frac{2}{\sqrt{x}+1}\Leftrightarrow\sqrt{x}+1=8\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)