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a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
a) Ta có: \(D=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\cdot\left(-\sqrt{2}\right)\)
\(=-2+\sqrt{6-2\sqrt{5}}\)
\(=-2+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)
\(=-2+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=-2+\left|\sqrt{5}-1\right|\)
\(=-2+\sqrt{5}-1\)(Vì \(\sqrt{5}>1\))
\(=-3+\sqrt{5}\)
b) Ta có: \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}\right)-\sqrt{75}\)
\(=2\sqrt{81}+4\sqrt{144}-5\sqrt{3}\)
\(=18+48-5\sqrt{3}\)
\(=66-5\sqrt{3}\)
c) Ta có: \(E=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=\sqrt{2}\cdot\left(5-3\right)\)
\(=2\sqrt{2}\)
d) Ta có: \(P=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\frac{3}{2}+2\cdot\sqrt{\frac{3}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}+\sqrt{\frac{3}{2}-2\cdot\sqrt{\frac{3}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}\)
\(=\sqrt{\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)^2}\)
\(=\left|\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right|+\left|\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right|\)
\(=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\)(Vì \(\sqrt{\frac{3}{2}}>\sqrt{\frac{1}{2}}>0\))
\(=2\sqrt{\frac{3}{2}}=\sqrt{4\cdot\frac{3}{2}}=\sqrt{6}\)
e) Ta có: \(M=-3\sqrt{50}+2\sqrt{98}-7\sqrt{72}\)
\(=\sqrt{2}\cdot\left(-3\cdot\sqrt{25}+2\cdot\sqrt{49}-7\cdot\sqrt{36}\right)\)
\(=\sqrt{2}\cdot\left(-15+14-42\right)\)
\(=-43\sqrt{2}\)
\(a,\sqrt{64a^2}+2a\left(a\ge0\right)\\ < =>\sqrt{8^2.a^2}+2a\\ < =>\sqrt{\left(8a\right)^2+2a}\\ < =>\left|8a\right|+2a\\ < =>8a+2a\\ < =>10a\left(TM\right)vìa\ge0\)
\(b,3\sqrt{9a^6}-6a^3\left(a\in R\right)\\ < =>3\sqrt{\left(3a^2\right)^2}-6a^3\\ < =>3\left|3a^3\right|-6a^3\\ \)
Nếu \(a\ge0\) thì giá trị của biểu thức là:
\(3.3a^2-6a^2\\ =9a^3-6a^3\\ =3a^3\)
Nếu a<0 thì giá trị của biểu thức là:
\(3\left(-3a^3\right)-6a^3=-9a^3\\ =-6a^3=-15a^3\)
\(c,\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}\left(a\ge3\right)\\ =\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}\\ =\left|a+3\right|+\left|a-3\right|\\ =a+3+a-3\\ =2a\)
1: ĐKXĐ: 2-3x>=0
=>x<=2/3
2: ĐKXĐ: -3x^2>=0
=>x^2<=0
=>x=0
3: ĐKXĐ: -2023x^3>=0
=>x^3<=0
=>x<=0
4: ĐKXĐ: -2(x-5)>=0
=>x-5<=0
=>x<=5
5: ĐKXĐ: -5/2-2x>=0
=>2-2x<0
=>2x>2
=>x>1
6: ĐKXĐ: (x^2+1)(3-2x)>=0
=>3-2x>=0
=>-2x>=-3
=>x<=3/2
7: ĐKXĐ: (-x^2-1)(3-x)>=0
=>(x^2+1)(x-3)>=0
=>x-3>=0
=>x>=3
\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)
d, \(D=\sqrt{3+2\sqrt{2}}=\sqrt{2+2.\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
e,\(E=\sqrt{8-2\sqrt{15}}=\sqrt{5-2.\sqrt{5}.\sqrt{3}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}\)
a,ĐKXĐ: \(\forall x\in R\)
\(\Rightarrow A=\left|a+3\right|+\left|a-3\right|\)\(=\left|-a-3\right|+\left|a-3\right|\)
Vì \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) *Dấu ''='' xảy ra\(\Leftrightarrow A.B\ge0\) *
\(\Rightarrow A\ge\left|-a-3+a-3\right|=6\)
Dấu ''='' xảy ra \(\Leftrightarrow\left(-a-3\right)\left(a-3\right)\ge0\Leftrightarrow\left(a+3\right)\left(a-3\right)\ge0\)
\(\Leftrightarrow-3\le a\le3\)
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