Question

bài 5 ạ, cảm ơn

Answer 1

1) Ta có: 

\(x=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\\ =\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-1\right)\\ =\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\\ =-\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)\\ =-\left(1-5\right)\\ =4\)

Thay x = 4 vào B ta có:

\(B=\dfrac{4+2}{4+\sqrt{4}+1}=\dfrac{6}{4+2+1}=\dfrac{6}{7}\) 

2) 

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

 3) 

\(P=\dfrac{A}{1-B}=\dfrac{2}{x+\sqrt{x}+1}:\left(1-\dfrac{x+2}{x+\sqrt{x}+1}\right)\\ =\dfrac{2}{x+\sqrt{x}+1}:\dfrac{x+\sqrt{x}+1-x-2}{x+\sqrt{x}+1}\\ =\dfrac{2}{x+\sqrt{x}+1}:\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\\ =\dfrac{2}{x+\sqrt{x}+1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{2}{\sqrt{x}-1}\) 

\(P\le1=>\dfrac{2}{\sqrt{x}-1}\le1\\ =>1-\dfrac{2}{\sqrt{x}-1}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\ge0\\\Leftrightarrow\left[{}\begin{matrix}x\ge9\\x< 1\end{matrix}\right.\) 

Kết hợp với đkxđ: `x>=9` hoặc `0<=x<1`

Answer 2

Bài VI

Gọi hai chữ số cần tìm là \(\overline{ab}\) \(\left(a;b\right)\in N\) và \(a\in\left\{0;1;2...9\right\};b\in\left\{0;1;...9\right\}\)

Theo đề bài ta có :

\(a+2b=12\left(1\right)\)

Khi đổi chỗ ta được :

\(\overline{ba}-\overline{ab}=10b+a-10a-b=9b-9a=27\)

\(\Leftrightarrow9\left(b-a\right)=27\)

\(\Leftrightarrow b-a=3\left(2\right)\)

\(\left(1\right)\&\left(2\right)\) ta được HPT

\(\left\{{}\begin{matrix}a+2b=12\\b-a=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b-3\\3b=15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=5\end{matrix}\right.\)

Vậy số ban đầu là \(25\)