Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
a) Ta có: \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
Để A>1 thì A-1>0
\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}-1>0\)
\(\Leftrightarrow\frac{\sqrt{x}-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{2}{\sqrt{x}-2}>0\)
mà 2>0
nên \(\sqrt{x}-2>0\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4(nhận)
Vậy: Khi x>4 thì A>1
B = ( 1/2^2 - 1) . ( 1/3^2 - 1) . ( 1/4^2 - 1) ... ( 1/98^2 - 1) . ( 1/99^2 - 1)
B = -3/2^2 . ( -8/3^2) . ( -15/4^2) .... ( -9603/98^2) . ( -9800/99^2)
Tích B gồm ( 99 - 2) : 1 + 1 = 98 ( số hạng), mỗi số hạng đều mang dấu âm nên khi tính ra kết quả tích B mang dấu dương
=> B = 3/2^2 . 8/3^2 . 15/4^2 ... 9603/98^2 . 9800/99^2
B = 1.3/2.2 . 2.4/3.3 . 3.5/4.4 ... 97.99/98.98 . 98.100/99.99
B = 1.2.3...97.98/2.3.4...98.99 . 3.4.5...99.100/2.3.4...98.99
B = 1/99 . 100/2
B = 1/99 . 50 = 50/99