Chứng minh các đẳng thức: 4 2 - 5 2 - 4 = 8
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10 tháng 9 2019
Ta có VT: \(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)=\(\frac{\sqrt{4}}{\sqrt{\left(2-\sqrt{5}\right)^2}}-\frac{\sqrt{4}}{\sqrt{\left(2+\sqrt{5}\right)^2}}\)
=\(\frac{2}{\left|2-\sqrt{5}\right|}-\frac{2}{\left|2+\sqrt{5}\right|}\)
=\(\frac{2}{\sqrt{5}-2}-\frac{2}{2+\sqrt{5}}\)
=\(\frac{2.\left(2+\sqrt{5}\right)-2.\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right).\left(2+\sqrt{5}\right)}\)
=\(2.\left(2+\sqrt{5}\right)-2.\left(\sqrt{5}-2\right)\)
=\(4+2\sqrt{5}-2\sqrt{5}+4\)
=8 (bằng VP)
17 tháng 12 2023
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
AT
7 tháng 7 2021
\(\sqrt{4x+8}+3\sqrt{x+2}=3+\dfrac{4}{5}\sqrt{25x+50}\left(x\ge-2\right)\)
\(\Rightarrow2\sqrt{x+2}+3\sqrt{x+2}-4\sqrt{x+2}=3\Rightarrow\sqrt{x+2}=3\Rightarrow x=7\)
\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
MN
8 tháng 8 2020
a) Sửa đề :
\(x^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
\(x^4=\left(a^4+3a^3b+3a^2b^2+ab^3\right)+\left(a^3b+3a^2b^2+3ab^3+b^4\right)\)
\(x^4=a\left(a^3+3a^2b+3ab^2+b^3\right)+b\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(x^4=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(x^4=\left(a+b\right)\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)
\(x^4=\left(a+b\right)\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)
\(x^4=\left(a+b\right)^2\left(a+2ab+b^2\right)\)
\(x^4=\left(a+b\right)^4\)
b) Sửa đề:
\(x^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)
\(x^5=\left(a^5+4a^4b+6a^3b^2+4a^2b^3+ab^4\right)+\left(a^4b+4a^3b^2+6a^2b+4ab^4+b^5\right)\)
\(x^5=a\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)+b\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)
\(x^5=\left(a+b\right)\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)
\(x^5=\left(a+b\right)\left[\left(a^4+3a^3b+3a^2b^2+ab^3\right)+\left(a^3b+3a^2b^2++3ab^3+b^4\right)\right]\)
\(x^5=\left(a+b\right)\left[a\left(a^3+3a^2b+3ab^2+b^3\right)+b\left(a^3+3a^2b+3ab^2+b^3\right)\right]\)
\(x^5=\left(a+b\right)^2\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(x^5=\left(a+b\right)^2\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)
\(x^5=\left(a+b\right)^2\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)
\(x^5=\left(a+b\right)^3\left(a^2+2ab+b^2\right)\)
\(x^5=\left(a+b\right)^5\)
Bạn có thể tự tóm tắt lại
2 tháng 6 2017
- \(\sqrt{\sqrt{5}^2-2.2\sqrt{5}+4}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\left(dpcm\right)\)
- \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{\sqrt{7}^2+2.4\sqrt{7}+16}-\sqrt{7}\)\(=\sqrt{\left(\sqrt{7}+4\right)^2}-\sqrt{7}=\sqrt{7}+4-\sqrt{7}=4\left(DPCM\right)\)
DN
0
Vế trái bằng vế phải nên đẳng thức được chứng mih