![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 99 x2 + 33+333+3333+...+333...33 (100 chữ số 3)
đặt B = 33+333+3333+...+3333....33 = (99+999+9999+....+999...99): 3
= (102-1+103-1+104-1+....+ 10100-1):3
= (102+103+104+....+ 10100-99):3
đặt C= 102+103+104+....+ 10100
Cx 10 = 103+104+....+ 10100+10101
Suy ra C x9 = 10101- 102
Bạn tính tiếp nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-6\sqrt{20}+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n=\left(1+\sqrt{3}+\sqrt{5}\right)\left(1+\sqrt{3}-\sqrt{5}\right)\)
\(n=\left(1+\sqrt{3}\right)^2-\sqrt{5}^2\)
\(n=1+2.\sqrt{3}.1+3-25\)
\(n=4-25+2\sqrt{3}\)
\(n=-21+2\sqrt{3}\)
![](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 \(1+3\sqrt{3}\).