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15 tháng 10 2020

\(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{\left(\sqrt{45}\right)^2+\left(\sqrt{3}\right)^2-2\cdot\sqrt{45}\cdot\sqrt{3}}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{\left(\sqrt{45}-\sqrt{3}\right)^2}-\sqrt{45}+\sqrt{18}\)

\(=\left|\sqrt{45}-\sqrt{3}\right|-\sqrt{45}+3\sqrt{2}\)

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

21 tháng 12 2018

a) \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}=\left|\sqrt{3}-3\right|+\sqrt{3-2\sqrt{3}+1}=3-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\left|\sqrt{3}-1\right|=3-\sqrt{3}+\sqrt{3}-1=2\)

b) \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}=\sqrt{48-2\sqrt{9.15}}-\sqrt{9.5}+\sqrt{9.2}=\sqrt{48-6\sqrt{15}}-3\sqrt{5}+3\sqrt{2}=\sqrt{3-2.\sqrt{3}.3\sqrt{5}+45}-3\sqrt{5}+3\sqrt{2}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.3\sqrt{5}+\left(3\sqrt{5}\right)^2}-3\sqrt{5}+3\sqrt{2}=\sqrt{\left(\sqrt{3}-3\sqrt{5}\right)^2}-3\sqrt{5}+3\sqrt{2}=\left|\sqrt{3}-3\sqrt{5}\right|-3\sqrt{5}+3\sqrt{2}=3\sqrt{5}-\sqrt{3}-3\sqrt{5}+3\sqrt{2}=3\sqrt{2}-\sqrt{3}\)

Ta có: \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{45-2\cdot\sqrt{45}\cdot\sqrt{3}+3}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{\left(\sqrt{45}-\sqrt{3}\right)^2}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{45}-\sqrt{3}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{18}-\sqrt{3}\)

10 tháng 7 2017

\(A=\sqrt{45-2\sqrt{135}+3}-3\sqrt{5}+3\sqrt{2}\\ =\sqrt{\left(3\sqrt{5}-\sqrt{3}\right)^2}-3\sqrt{5}+3\sqrt{2}\\ =3\sqrt{5}-\sqrt{3}-3\sqrt{5}+3\sqrt{2}\\ =3\sqrt{2}-\sqrt{3}\)

10 tháng 7 2017

\(B=\dfrac{\sqrt{5}.\sqrt{2}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}-\dfrac{6\left(2+\sqrt{10}\right)}{10-4}-2\sqrt{10}\\ =\sqrt{10}-\dfrac{12+6\sqrt{10}}{6}-2\sqrt{10}\\ =\sqrt{10}-2-\sqrt{10}-2\sqrt{10}\\ =-2-2\sqrt{10}\)

16 tháng 12 2022

a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)

b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)

c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)

d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)

f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)

19 tháng 4 2020

\(\sqrt{48-2.3\sqrt{5}.\sqrt{3}}-\sqrt{45}+\sqrt{18}=\sqrt{\left(3\sqrt{5}-\sqrt{3}\right)^2}-3\sqrt{5}+3\sqrt{2}\)

\(=|3\sqrt{5}-\sqrt{3}|-3\sqrt{5}+3\sqrt{2}=3\sqrt{5}-\sqrt{3}-3\sqrt{5}+3\sqrt{2}=3\sqrt{2}-\sqrt{3}\)

học tốt

Bài 2:

a: \(=\sqrt{5}-2\)

b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)

c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)

d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)

e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)

\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)

f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)

\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)

\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)

15 tháng 10 2023

a: ĐKXĐ: x-5>=0

=>x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x-1>=0

=>x>=1

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)

=>\(-2\sqrt{x-1}=4\)

=>\(\sqrt{x-1}=-2\)(vô lý)

Vậy: Phương trình vô nghiệm

c: ĐKXĐ: x-2>=0

=>x>=2

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)

=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

=>\(\sqrt{x-2}\left(\dfrac{1}{3}-2+\dfrac{2}{3}\right)=-4\)

=>\(-\sqrt{x-2}=-4\)

=>x-2=16

=>x=18(nhận)

d: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)

=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)

=>\(4\sqrt{x+3}=0\)

=>x+3=0

=>x=-3(nhận)

15 tháng 10 2023

a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)

\(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(2\sqrt{x-5}=4\)

\(\sqrt{x-5}=2\)

\(\left|x-5\right|=4\)

=> \(x-5=\pm4\)

\(x=\pm4+5\)

\(x=9;x=1\)

Vậy x=9; x=1