Rut gon:
\(\sqrt{\dfrac{2}{5}}\)+\(\sqrt{\dfrac{5}{2}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 1 2019
\(P=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}+\sqrt{3+\sqrt{5}}\right)}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}\)
\(=\dfrac{3\sqrt{2}+\sqrt{10}}{4+\sqrt{6+2\sqrt{5}}}+\dfrac{3\sqrt{2}-\sqrt{10}}{4-\sqrt{6-2\sqrt{5}}}\)
\(=\dfrac{3\sqrt{2}+\sqrt{10}}{5+\sqrt{5}}+\dfrac{3\sqrt{2}-\sqrt{10}}{5-\sqrt{5}}\)
\(=\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(5-\sqrt{5}\right)+\left(3\sqrt{2}-\sqrt{10}\right)\left(5+\sqrt{5}\right)}{20}\)
\(=\dfrac{15\sqrt{2}-3\sqrt{10}+5\sqrt{10}-5\sqrt{2}+15\sqrt{2}+3\sqrt{10}-5\sqrt{10}-5\sqrt{2}}{20}\)
\(=\dfrac{30\sqrt{2}-10\sqrt{2}}{20}=\dfrac{20\sqrt{2}}{20}=\sqrt{2}\)
\(\)
26 tháng 5 2021
\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)
\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)^3}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{-\left(\sqrt{x}+2\right)^2}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1 2019
\(P=\dfrac{\left(\sqrt{a+1}+1\right)\left(\sqrt{a+1}+2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}+\dfrac{2\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}-\dfrac{2+5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{a+3+3\sqrt{a+1}}{a-3}+\dfrac{2a+2-4\sqrt{a+1}}{a-3}-\dfrac{2+5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{a+3+3\sqrt{a+1}+2a+2-4\sqrt{a+1}-2-5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{3a+3-6\sqrt{a+1}}{a-3}\)
Có thể dừng ở đây hoặc nếu thích thì làm tiếp như sau (chưa chắc gọn hơn):
\(P=\dfrac{3\left(a+1\right)-6\sqrt{a+1}}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}=\dfrac{3\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}\)
\(P=\dfrac{3\sqrt{a+1}}{\sqrt{a+1}-2}\)
Y
21 tháng 5 2023
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{\sqrt{x}+2}\left(dkxd:x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}+2}{x+4}\right)\)
\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)
\(=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
Vậy \(B=\dfrac{1}{\sqrt{x}-2}\)
19 tháng 8 2023
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)
16 tháng 10 2018
Cho \(5\sqrt{x}7\) mk viet nham
Sua lai thanh \(5\sqrt{x}-7\)
19 tháng 10 2022
a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)
=>10 căn x+5-5 chia hết cho 2 căn x+1
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
hay \(x\in\varnothing\)
13 tháng 10 2022
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}+\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{2}\right)}{-\sqrt{x}}\)
2 tháng 1 2023
a: \(P=\dfrac{\left(\sqrt{x^2+1}\right)^2+5\sqrt{x^2+1}+6}{\sqrt{x^2+1}+3}+\dfrac{\left(\sqrt{x^2+1}^2\right)+7\sqrt{x^2+1}+12}{\sqrt{x^2+1}+4}\)
\(=\sqrt{x^2+1}+2+\sqrt{x^2+1}+3\)
\(=2\sqrt{x^2+1}+5\)
b: Để P=11 thì \(2\sqrt{x^2+1}=11-5=6\)
=>căn (x^2+1)=3
=>x^2+1=9
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
25 tháng 12 2018
XEM CÓ SAI ĐỀ BÀI KHÔNG, MK RÚT GỌN RA TO LẮM
12 tháng 12 2022
\(=\dfrac{x+5\sqrt{x}+6-x+5\sqrt{x}-6}{\left(\sqrt{x}+3\right)^2\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{\sqrt{x}}\)
\(=\dfrac{10\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{10}{\sqrt{x}+3}\)
LV
1
18 tháng 8 2018
\(P=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
\(\sqrt{\dfrac{2}{5}}\) + \(\sqrt{\dfrac{5}{2}}\) = \(\dfrac{\sqrt{2}\times\sqrt{2}}{\sqrt{10}}\)+ \(\dfrac{\sqrt{5}\times\sqrt{5}}{\sqrt{10}}\) = \(\dfrac{7}{\sqrt{10}}\)= \(\dfrac{7\sqrt{10}}{10}\)
Lời giải:
\(\sqrt{\frac{2}{5}}+\sqrt{\frac{5}{2}}=\frac{\sqrt{2}.\sqrt{2}+\sqrt{5}.\sqrt{5}}{\sqrt{5}.\sqrt{2}}=\frac{7}{\sqrt{10}}=\frac{7\sqrt{10}}{10}\)