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BH
15 tháng 8 2018
\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)
\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)
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)
VH
6 tháng 11 2019
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-3+\sqrt{2}\)
\(=\sqrt{5}-2-3+\sqrt{2}=\sqrt{5}+\sqrt{2}-5\)
2 tháng 9 2018
\(\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6+\sqrt{2}}\right)}=2\)
=2.
29 tháng 6 2020
\(\frac{2x}{x-2}-\frac{3x+10}{x^2-4}=\frac{x}{x+2}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2x}{x-2}-\frac{3x+10}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x+2}=0\)
\(\Leftrightarrow\frac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3x+10}{\left(x-2\right)\left(x+2\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}-\frac{3x+10}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x^2+4x-3x-10-x^2+2x}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{x^2+3x-10}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{x^2+5x-2x-10}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
=> x+5=0
<=> x=-5(tmđk)
Vậy x=-5 là nghiệm của phương trình
29 tháng 6 2020
\(\frac{2x}{x-2}-\frac{3x+10}{x^2-4}=\frac{x}{x+2}\) ( đkxđ : \(x\ne\pm2\))
\(\Leftrightarrow\frac{2x}{x-2}-\frac{3x+10}{\left(x+2\right)\left(x-2\right)}=\frac{x}{x+2}\)
\(\Leftrightarrow\frac{2x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3x+10}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow2x^2+4x-3x-10=x^2-2x\)
\(\Leftrightarrow2x^2+4x-3x-10-x^2+2x=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
\(x\ne\pm2\)=> x = -5
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2020
\(A>0\)
\(A^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{5}+1\)
24 tháng 8 2019
1, △ABC△ABC vuông có ˆA=900A^=900 , ˆB=600B^=600 và b = 10 thì độ dài a là :
A. a = 15√3153
B. a = 10√3103
C. a = 20√332033
D. a = 20√3203
2, △ABC△ABC vuông có ˆA=900,ˆC=600A^=900,C^=600 và b = thì độ dài b' là :
A. b' = 8
B. b' = 6
C. b' = 6√363
D. b' = 3√3
Ta có:
\(2\sqrt{50}+\sqrt{36}-10\sqrt{2}\\ =10\sqrt{2}+6-10\sqrt{2}\\ =6\)