Rút gọn biểu thức sau:
\(A=\left(\sqrt{22}+7\sqrt{2}\right)\sqrt{30-7\sqrt{11}}\)
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
21 tháng 10 2023
1:
a: \(\sqrt{25}+\sqrt{49}=5+7=12\)
b: \(\sqrt{121}-\sqrt{81}=11-9=2\)
2: x>-2
=>2x>-4
=>2x+1>-3
=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa
3:
a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}=-1\)
b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)
\(=21-14\sqrt{2}+14\sqrt{2}=21\)
c:
\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)
16 tháng 6 2017
a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)
= \(6-\sqrt{15}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)
c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)
= \(7\)
d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)
HH
23 tháng 4 2017
a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15
b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10
c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7
d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22
21 tháng 7 2023
`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.
`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.
`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.
`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.
`= 7`.
`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`
`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`
`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`
`= 33 - 11 = 22`.
28 tháng 5 2022
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
LT
25 tháng 6 2021
\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)
\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)
\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)
\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\)
25 tháng 6 2021
Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(=\sqrt{7}-2+8-2\sqrt{7}\)
\(=6-\sqrt{7}\)
4 tháng 10 2021
\(\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{4-\sqrt{7}}}{\sqrt{4+\sqrt{7}}}\)
\(=\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{7}-1}{\sqrt{7}+1}\)
\(=\dfrac{\left(\sqrt{7}+1\right)^2\cdot\left(\sqrt{7}-1\right)}{\sqrt{7}+1}\cdot\dfrac{1}{2}\)
\(=\dfrac{6}{2}=3\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2021
\(=\dfrac{\left(8+2\sqrt{7}\right)\sqrt{8-2\sqrt{7}}}{2\sqrt{8+2\sqrt{7}}}=\dfrac{\left(\sqrt{7}+1\right)^2\sqrt{\left(\sqrt{7}-1\right)^2}}{2\sqrt{\left(\sqrt{7}+1\right)^2}}\)
\(=\dfrac{\left(\sqrt{7}+1\right)^2\left(\sqrt{7}-1\right)}{2\left(\sqrt{7}+1\right)}=\dfrac{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}{2}\)
\(=\dfrac{7-1}{2}=3\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1
\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)
\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)
11 tháng 1
\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)
\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)
\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)
b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)
\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)
\(=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{2}}=a^2\)
8 tháng 6 2021
Đk:\(a>2\)
\(\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
Đặt \(b=\sqrt{a-2}\Leftrightarrow a=b^2+2\)
Biểu thức \(\Leftrightarrow\dfrac{b+2}{3}\left(\dfrac{b}{3+b}+\dfrac{b^2+2+7}{11-b^2-2}\right):\left(\dfrac{3b+1}{b^2-3b}-\dfrac{1}{b}\right)\)
\(=\dfrac{b+2}{3}\left[\dfrac{b}{3+b}-\dfrac{b^2+9}{b^2-9}\right]:\left[\dfrac{3b+1}{b\left(b-3\right)}-\dfrac{b-3}{b\left(b-3\right)}\right]\)
\(=\dfrac{b+2}{3}.\dfrac{b\left(b-3\right)-b^2-9}{\left(b-3\right)\left(3+b\right)}:\dfrac{3b+1-\left(b-3\right)}{b\left(b-3\right)}\)
\(=\dfrac{b+2}{3}.\dfrac{-3\left(b+3\right)}{\left(b-3\right)\left(3+b\right)}.\dfrac{b\left(b-3\right)}{2\left(b+2\right)}\)
\(=-\dfrac{b}{2}\)
\(=\dfrac{\sqrt{a-2}}{-2}\)