Câu hỏi của Thơ Nụ =))

Question

bài 5 ạ, cảm ơn

HT.Phong (9A5) · Accepted Answer

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`Câu trả lời: 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`

Nguyễn Đức Trí · Answer

Bài VIGọi 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.$$\Leftrightarrow\left\{{}\begin{matrix}a=2\b=5\end{matrix}\right.$Vậy số ban đầu là $25$Câu trả lời: Bài VIGọi hai chữ số cần tìm là $\overline{ab}$ $\left(a;b\right)\in N$ và $a\in\left\{0;1;2...9\right\};b\in\left\{0;1;...9\right\}$Theo đề bài ta có :$a+2b=12\left(1\right)$Khi đổi chỗ ta được :$\overline{ba}-\overline{ab}=10b+a-10a-b=9b-9a=27$$\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 $25$

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)