A 7ab. \(\sqrt{\dfrac{36a^4}{49b^2}}\) với b > 0
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
30 tháng 6 2021
`a)sqrt{(sqrt7-4)^2}+sqrt7`
`=|sqrt7-4|+sqrt7`
`=4-sqrt7+sqrt7=4`
`b)\sqrt{81a}-sqrt{144a}+sqrt{36a}(a>=0)`
`=9sqrta-12sqrta+6sqrta=3sqrta`
30 tháng 6 2021
a) Ta có: \(\sqrt{\left(\sqrt{7}-4\right)^2}+\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
b) Ta có: \(\sqrt{81a}-\sqrt{144a}+\sqrt{36a}\)
\(=9\sqrt{a}-12\sqrt{a}+6\sqrt{a}\)
\(=3\sqrt{a}\)
3 tháng 9 2022
a: \(=-xy\cdot\dfrac{\sqrt{xy}}{x}=-y\sqrt{yx}\)
b: \(=\sqrt{\dfrac{-105x^3}{35^2}}=\sqrt{-105x}\cdot\dfrac{x}{35}\)
c: \(=\sqrt{\dfrac{5a^3b}{49b^2}}=\sqrt{5ab}\cdot\dfrac{a}{7b}\)
d: \(=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3}\cdot\sqrt{xy}\)
6 tháng 10 2022
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
27 tháng 7 2017
a, \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)
\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)
\(=-\sqrt{3}\)
b, \(\sqrt{81a}-\sqrt{36a}+\sqrt{144a}\)
\(=9\sqrt{a}-6\sqrt{a}+12\sqrt{a}\)
\(=15\sqrt{a}\)
c, \(\dfrac{4}{\sqrt{5}-2}-\dfrac{4}{\sqrt{5}+2}\)
\(=\dfrac{4\sqrt{5}+8-4\sqrt{5}+8}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{16}{5-4}=16\)
d, \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{ab}\)
27 tháng 7 2017
Nguyễn Huy Tú anh sinh năm 2004 là lên lớp 8 mà sao lại tl được bài lớp 9
15 tháng 11 2021
\(PT\Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\sqrt{\dfrac{\dfrac{2x+1}{2}}{2}}=\dfrac{3}{2}\\ \Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\dfrac{1}{2}\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow2\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow\sqrt{2x+1}=\dfrac{3}{4}\\ \Leftrightarrow2x+1=\dfrac{9}{16}\\ \Leftrightarrow2x=-\dfrac{7}{16}\\ \Leftrightarrow x=-\dfrac{7}{32}\\ \Leftrightarrow a=-\dfrac{7}{32}\\ \Leftrightarrow1-36a=1+36\cdot\dfrac{7}{32}=...\)
26 tháng 8 2017
bài 1) a) \(xy\sqrt{\dfrac{x}{y}}=x\sqrt{y}\sqrt{y}\dfrac{\sqrt{x}}{\sqrt{y}}=x\sqrt{x}\sqrt{y}=\left(\sqrt{x}\right)^3\sqrt{y}\)
b) \(\sqrt{\dfrac{5a^3}{49b}}=\dfrac{\sqrt{5a^3}}{\sqrt{49b}}=\dfrac{\sqrt{5a^3}}{7\sqrt{b}}=\dfrac{\sqrt{5a^3}.\sqrt{b}}{7\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{5a^3b}}{7b}\)
bài 2) a) \(\dfrac{\sqrt{3}-3}{1-\sqrt{3}}=\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=\sqrt{3}\)
b) \(\dfrac{5-\sqrt{15}}{\sqrt{3}-\sqrt{5}}=\dfrac{-\sqrt{5}\left(\sqrt{3}-\sqrt{5}\right)}{\sqrt{3}-\sqrt{5}}=-\sqrt{5}\)
c) \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
TM
5 tháng 7 2023
(a) \(\sqrt{\dfrac{a^2}{25+10b+b^2}}=\sqrt{\dfrac{a^2}{\left(5+b\right)^2}}=\dfrac{\sqrt{a^2}}{\sqrt{\left(5+b\right)^2}}\)
\(=\dfrac{\left|a\right|}{\left|5+b\right|}=\dfrac{-a}{b+5}\) (do \(a< 0,b>0\Rightarrow b+5>0\))
(b) \(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right)\sqrt{\dfrac{\left(ab\right)^2}{\left(a-b\right)^2}}=\left(a-b\right)\cdot\dfrac{\sqrt{\left(ab\right)^2}}{\sqrt{\left(a-b\right)^2}}\)
\(=\left(a-b\right)\cdot\dfrac{\left|ab\right|}{\left|a-b\right|}\).
(c) \(\dfrac{x+4\sqrt{x}+4}{2+\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}=\sqrt{x}+2.\)
25 tháng 10 2018
Bài 1
a) √81a - √36a - √144a = 9√a - 6√a - 12√a = -9√a
b) √75 - √48 - √300 = 5√3 - 4√3 - 10√3 = -9√3
Bài 2
a) √2x-3 = 7
⇒ 2x-3 = 49 ⇔ 2x = 52 ⇔ x =26
c) √16x - √9x = 2
⇔ 4√x - 3√x = 2 ⇔ √x = 2 ⇔ x = 4
Bài 3
a) √(2-√5)2 = l 2-√5 l = √5-2
b) (a - 3)2 + (a - 9)
= a2 - 6a + 9 + a - 9 = a2 - 5a
c) A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
=\(\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{-3\sqrt{x}-3}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\left(\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\dfrac{-3\sqrt{x}+9}{x-9}\)
\(7ab\cdot\sqrt{\dfrac{36a^4}{49b^2}}\\ =>7ab\cdot\sqrt{\left(\dfrac{6a^2}{7b}\right)^2}\\ =>7ab\cdot\dfrac{6a^2}{7b}\\ =>\dfrac{7ab\cdot6a^2}{7b}\\ =>6a^3\)
\(A=7ab.\sqrt{\dfrac{36a^4}{49b^2}}\)
\(=7ab.\dfrac{6a^2}{\left|7b\right|}\)
\(=7ab.\dfrac{6a^2}{7b}\left(vib>0\right)\)
\(=6a^3\)