Rút gọn \(\sqrt{8+2\sqrt15}-\sqrt{8-2\sqrt15}\)
\(\sqrt{(5+2\sqrt6)}+\sqrt{8-2\sqrt15}\)
\(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}-\dfrac{5}{\sqrt3-2\sqrt2}-\dfrac{5}{\sqrt3+\sqrt8}\)
K
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Những câu hỏi liên quan
14 tháng 8 2019
\(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=1-\sqrt{3}-\sqrt{3}-2\)
\(=-2\sqrt{3}-1\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+4-2\sqrt{3}\)
\(=6-3\sqrt{3}\)
MS
2
28 tháng 10 2023
\(\left(3\sqrt{12}-4\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-2\sqrt{5}\)
\(=\left(6\sqrt{3}-4\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-2\sqrt{5}\)
\(=6+3\sqrt{5}-2\sqrt{5}=6+\sqrt{5}\)
H
27 tháng 10 2023
Bài `1`
\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)
27 tháng 10 2023
2:
a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
b: B=5
=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)
=>\(5\sqrt{x}+15=\sqrt{x}+8\)
=>\(4\sqrt{x}=-7\)(loại)
Vậy: \(x\in\varnothing\)
5 tháng 8 2020
1/ \(A=\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}-\sqrt{3}>0\))
\(B=\sqrt{6+2\sqrt{5}}-\sqrt{13}+\sqrt{48}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{13}+4\sqrt{3}=\left|\sqrt{5}+1\right|-\sqrt{13}+4\sqrt{3}=\sqrt{5}+1+\sqrt{13}+4\sqrt{5}\)
2/Ta có :
\(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{3\sqrt{2}}{3\sqrt{3}-3}-\frac{5\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{3\sqrt{2}}{3\left(\sqrt{3}-1\right)}-\frac{5\sqrt{6}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}-1\right)}\right).\frac{1}{\sqrt{6}}\)
\(=\frac{3\sqrt{2}-15\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)
\(=\frac{-12\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)
\(=\frac{-7+\sqrt{3}}{6}\)
Vậy...
5 tháng 8 2020
Bài 1:
Ta có: \(A=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}-2\cdot\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left|\sqrt{5}+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|-2\cdot\left|\sqrt{5}-1\right|\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2\)
=2
Vậy: A=2
Bài 2: Sửa đề: Chứng minh \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}=\frac{-7+\sqrt{3}}{6}\)
Ta có: \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{9\sqrt{2}}{3\left(\sqrt{27}-3\right)}-\frac{\sqrt{150}\left(\sqrt{27}-3\right)}{3\cdot\left(\sqrt{27}-3\right)}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{9\sqrt{2}-45\sqrt{2}+3\sqrt{150}}{9\left(\sqrt{3}-1\right)}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-36\sqrt{2}+3\sqrt{150}}{9\sqrt{6}\cdot\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{54}\cdot\left(5-4\sqrt{3}\right)}{\sqrt{486}\cdot\left(\sqrt{3}-1\right)}\)
\(=\frac{5-4\sqrt{3}}{3\sqrt{3}-3}\)
\(=\frac{-7+\sqrt{3}}{6}\)(đpcm)
PK
1
2 tháng 9 2018
\(\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6+\sqrt{2}}\right)}=2\)
=2.
MS
1
H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
\(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}+\dfrac{3}{\sqrt{3}}\)
\(=\left|\sqrt{3}-\sqrt{5}\right|-\left|1-\sqrt{5}\right|+\dfrac{\left(\sqrt{3}\right)^2}{\sqrt{3}}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}-1\right)+\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}+1+\sqrt{3}\)
\(=1\)
NT
1
13 tháng 8 2020
Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=21\cdot\left[2+\sqrt{3}+3-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right]-6\cdot\left[2-\sqrt{3}+3+\sqrt{5}+2\cdot\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{\left(4+2\sqrt{3}\right)\left(6-2\sqrt{5}\right)}\right)-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\sqrt{\left(4-2\sqrt{3}\right)\left(6+2\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left[5+\sqrt{3}-\sqrt{5}+\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\right]-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\left(\sqrt{3}-1\right)\left(\sqrt{5}+1\right)\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{15}-\sqrt{3}+\sqrt{5}-1\right)-6\cdot\left(5-\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt{3}-\sqrt{5}-1\right)-15\sqrt{15}\)
\(=21\cdot\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)
\(=60\)
\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)
\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)