kb với tui đi

cậu kb tui ik

Xin phép sửa đề : đoạn cuối n -> m  

Bài làm :

PT <=> \(\sqrt{x-9}+3+m\left(\sqrt{x-9}+1\right)=x+\frac{3m+1}{2}\)

Đặt \(t=\sqrt{x-9}\left(t\ge0\right)\)

PT trở thành :

\(t+3+m\left(t+1\right)=t^2+9+\frac{3m+1}{2}\)

\(\Leftrightarrow2t^2-2\left(m+1\right)t+m+13=0\left(1\right)\)

PT ban đầu có nghiệm \(x_1< 10< x_2\)

<=> (1) có nghiệm \(0\le t_1< 1< t_2\Leftrightarrow\hept{\begin{cases}\Delta'>0\\\left(t_1-1\right)\left(t_2-1\right)< 0\\t_1+t_2>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(m+1\right)^2-2\left(m+13\right)>0\\\frac{m+13}{2}-m-1+1< 0\\m+1>0\end{cases}\Leftrightarrow m>13}\)

taijsao delta' nhỏ hơn 0 bạn

c) \(\frac{x^2-2}{x+\sqrt{2}}=\frac{x^2-\left(\sqrt{2}\right)^2}{x+\sqrt{2}}=\frac{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}{x+\sqrt{2}}=x-\sqrt{2}\)

1×9999÷98767×0299999=30371,37911

nhớ tít nha

ĐK: \(x\ge0,x\ne9\).

\(P=\frac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{3-\sqrt{x}}\)

\(=\frac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}-10+x-6\sqrt{x}+9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

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

\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)

Tại \(x=1-\frac{\sqrt{3}}{2}\): \(\sqrt{x}=\sqrt{1-\frac{\sqrt{3}}{2}}=\frac{1}{2}\sqrt{4-2\sqrt{3}}=\frac{1}{2}\sqrt{\left(\sqrt{3}-1\right)^2}=\frac{1}{2}\left(\sqrt{3}-1\right)\)

\(P=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+4}=\frac{\sqrt{3}-3}{\sqrt{3}+3}=-2+\sqrt{3}\).

\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)

Dễ thấy \(P< 1\).

\(\sqrt{x}+2\ge2\Rightarrow\frac{3}{\sqrt{x}+2}\le\frac{3}{2}\Rightarrow P\ge1-\frac{3}{2}=-\frac{1}{2}\)

Suy ra \(-\frac{1}{2}\le P< 1\)do đó \(P\)chỉ có thể nhận \(1\)giá trị nguyên duy nhất là \(P=0\).

Với \(P=0\Rightarrow x=1\). 

Do đó ta có đpcm.