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a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{2}\)
Biến đổi Vế trái ta có :
\(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}-\frac{3\sqrt{24}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(\left(\frac{\sqrt{3}}{\sqrt{2}}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{1}{\sqrt{6}}-2\sqrt{6}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{1}{2}-2=-1,5=VP\) ( ĐPCM)
Biến đổi vế trái :
\(VT=\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{2}.\sqrt{2}.\sqrt{3}-\sqrt{6}}{\sqrt{2^2.2}-2}-\frac{\sqrt{6^2.6}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{2}.\sqrt{6}-\sqrt{6}}{2\sqrt{2}-2}-\frac{6\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}.\left(\sqrt{2}-1\right)}{2.\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}\)
\(=\sqrt{6}.\left(\frac{1}{2}-2\right).\frac{1}{\sqrt{6}}=-\frac{3}{2}=-1,5=VP\left(đpcm\right)\)
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
giả sử 2 vế bằng nhau, nhân tích chéo, rồi được 2 vế = nhau là kết luận thỏa mãn
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1=vp\)
Chứng minh các bất đẳng thức:
a) căn 6 - căn 2 >1
b) căn 5 - căn 3>1/2
c) căn 7 - căn 6 < căn 6 - căn 5
a)\(\frac{3.\sqrt{6}}{2}+\frac{2.\sqrt{2}}{\sqrt{3}}-\frac{4.\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{2}.\sqrt{3}}{\sqrt{3}.\sqrt{3}}-\frac{4.\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{2\sqrt{6}}{3}-\frac{\sqrt{6}}{2}=\frac{4\sqrt{6}-3\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\)
--> dpcm
b) \(\left(\frac{-\sqrt{7}.\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}.\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right).\frac{\sqrt{7}-\sqrt{5}}{1}\)
=\(\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(7-5\right)\)
=-1.2
=-2
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}-\frac{2b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+b\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{4b\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2-4b\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\right)-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{2\sqrt{a}.2\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{a}\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{a}\sqrt{b}\left(1-\sqrt{b}-b\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{2\sqrt{a}\sqrt{b}\left(1-\sqrt{b}-b\right)}{a-b}\)
Đề sai???Phân số thứ 3 nghi là a-b chứ ko phải căn a - căn b????????
Biến đổi vế trái: