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23 tháng 4 2017

ĐS: a) 3√5;35;

b) 9√22;922;

c) 15√2−√5;152−5;

d) 17√25.


19 tháng 6 2017

a) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\) = \(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\) = \(-\sqrt{3}\)

b) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\) = \(7\sqrt{2}-6\sqrt{2}+\sqrt{2}\) = \(2\sqrt{2}\)

c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\) = \(3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\) = \(6\sqrt{a}\)

d) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\) = \(4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)

= \(4\sqrt{b}-5\sqrt{10b}\)

20 tháng 7 2023

2) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}\)

\(=\left(7-6+1\right)\sqrt{2}\)

\(=2\sqrt{2}\)

3) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)

\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)

\(=\left(3-4+7\right)\sqrt{a}\)

\(=6\sqrt{a}\)

4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)

\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)

\(=4\sqrt{b}-5\sqrt{10b}\)

20 tháng 7 2023

Gấp nha 

31 tháng 7 2021

a) \(A=\sqrt{9a}-\sqrt{16a}-\sqrt{49a}=3\sqrt{a}-4\sqrt{a}-7\sqrt{a}=-8\sqrt{a}\)

b) \(B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)

\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)

\(=2+\sqrt{3}+\sqrt{2}+1-\sqrt{3}-\sqrt{2}=3\)

30 tháng 11 2021

\(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}=4\sqrt{b}+2.2\sqrt{10b}-3.3\sqrt{10b}=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}=4\sqrt{b}-5\sqrt{10b}\)

30 tháng 11 2021

Có mấy câu nữa trả lời giúp em ạ

b: B=căn 49a^2+3a

=|7a|+3a

=7a+3a(a>=0)

=10a

c: C=căn16a^4+6a^2

=4a^2+6a^2

=10a^2

d: \(D=3\cdot3\cdot\sqrt{a^6}-6a^3=6\cdot\left|a^3\right|-6a^3\)

TH1: a>=0

D=6a^3-6a^3=0

TH2: a<0

D=-6a^3-6a^3=-12a^3

e: \(E=3\sqrt{9a^6}-6a^3\)

\(=3\cdot\sqrt{\left(3a^3\right)^2}-6a^3\)

=3*3a^3-6a^3(a>=0)

=3a^3

f: \(F=\sqrt{16a^{10}}+6a^5\)

\(=\sqrt{\left(4a^5\right)^2}+6a^5\)

=-4a^5+6a^5(a<=0)

=2a^5

8 tháng 6 2016

\(A=\sqrt{4^2b}+2\sqrt{2^2\cdot10b}-3\sqrt{3^2\cdot10b}=4\sqrt{b}+4\sqrt{10}\cdot\sqrt{b}-9\sqrt{10}\cdot\sqrt{b}\)

\(=4\sqrt{b}-5\sqrt{10}\sqrt{b}=\left(4-5\sqrt{10}\right)\sqrt{b}\)

8 tháng 6 2016

Rut gon A = √16b+2√40b−3√90bva`b≥0

 A=√42b+2√22·10b−3√32·10b=4√b+4√10·√b−9√10·√b

=4√b−5√10√b=(4−5√10)√b

21 tháng 7 2023

4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)

\(=\sqrt{4^2\cdot b}+2\sqrt{2^2\cdot10b}-3\sqrt{3^2\cdot10b}\)

\(=4\sqrt{b}+2\cdot2\sqrt{10b}-3\cdot3\sqrt{10b}\)

\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)

\(=4\sqrt{b}+\left(4\sqrt{10b}-9\sqrt{10b}\right)\)

\(=4\sqrt{b}-5\sqrt{10b}\)

`a, sqrt(16b) + 2 sqrt(40b) - 3 sqrt(90b)`

`= 4sqrtb + 2sqrt(8.5b) - 3 sqrt(9.10b)`

`= 4 sqrt b + 4sqrt(10b) - 9 sqrt(10b)`

`= 4sqrtb-5sqrt(10b)`.

\(=7\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}=7\sqrt{b}-5\sqrt{10b}\)

17 tháng 12 2020

a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)

\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)

\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)

c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)

d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)

\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)

e, Ghi đúng đề.

\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)

AH
Akai Haruma
Giáo viên
20 tháng 7 2020

Lời giải:

a)

$\sqrt{98}-\sqrt{72}+0.5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}$

$=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}$

b)

$\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}$

$=4\sqrt{a}+4\sqrt{10}.\sqrt{a}-9\sqrt{10}.\sqrt{a}$

$=(4+4\sqrt{10}-9\sqrt{10})\sqrt{a}=(4-5\sqrt{10}).\sqrt{a}$

c)

$(2\sqrt{3}+\sqrt{5})\sqrt{3}-\sqrt{60}=2.3+\sqrt{15}-2\sqrt{15}$

$=6-\sqrt{15}$

d)

$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$

$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$

$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$

$=3.11+\sqrt{2}(12-3\sqrt{11})-11$

$=22+\sqrt{2}(12-3\sqrt{11})$