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8 tháng 7 2021
\(-7xy\sqrt{\frac{16}{xy}}\)
\(-7xy\frac{4\sqrt{xy}}{xy}\)
\(-28\sqrt{xy}\)
ND
3
20 tháng 6 2021
\(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\left(đk:x\ge2;y\ge3;z\ge5\right)\)
\(< =>\left(x-2\right)-2\sqrt{x-2}+1+\left(y-3\right)-4\sqrt{y-3}+4+\left(z-5\right)-6\sqrt{z-5}+9=0\)
\(< =>\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Do \(\left(\sqrt{x-2}-1\right)^2\ge0;\left(\sqrt{y-3}-2\right)^2\ge0;\left(\sqrt{z-5}-3\right)^2\ge0\)
Cộng theo vế ta được \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2\ge0\)
Mà \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra khi và chỉ khi x = 3 ; y = 7 ; z = 14 ( tmđk )
Vậy ...
NT
2
TV
http://lazi.vn/uploads/edu/exercise/1510324551_13.jpg
0
LA
15 tháng 12 2017
\(A=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)
\(A=\sqrt{9.3}-2\sqrt{3.4}-\sqrt{25.3}\)
\(A=3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\)
\(A=-6\sqrt{3}\)
\(B=\frac{1}{3+\sqrt{7}}+\frac{1}{3-\sqrt{7}}\)
\(B=\frac{3-\sqrt{7}+3\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(B=\frac{6}{9-7}=3\)
AO
15 tháng 12 2017
\(A=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)
\(=\sqrt{3^2.3}-2.\sqrt{2^2.3}-\sqrt{5^2.3}\)
\(=3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\)
\(=-6\sqrt{3}\)
vậy \(A=-6\sqrt{3}\)
\(B=\frac{1}{3+\sqrt{7}}+\frac{1}{3-\sqrt{7}}\)
\(B=\frac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(B=\frac{3-\sqrt{7}+3+\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(B=\frac{6}{9-7}\)
\(B=\frac{6}{2}\)
\(B=3\)
vậy \(B=3\)
LH
![[âIMG]](https://lazi.vn/uploads/edu/exercise/1505311489_8.jpg)
3
HN
1
NT
giúp mình với nha - Mình cảm ơn!!
2
TT
2
P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
P = \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
P = \(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+\sqrt{25}}=\sqrt{9}=3\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+2\sqrt{3}+\sqrt{\left(16-8\sqrt{2}+2\right)}}}}\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
Q = \(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{3}-2}\)
Q = \(\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
D = \(\sqrt{2+\sqrt{3}+\sqrt{4-2\sqrt{3}-\sqrt{21-12\sqrt{3}}}}\)
D = \(\sqrt{2+\sqrt{3}+\sqrt{4-2\sqrt{3}-\sqrt{\left(2\sqrt{3}-3\right)^2}}}\)
D = \(\sqrt{2+\sqrt{3}+\sqrt{4-2\sqrt{3}-2\sqrt{3}+3}}\)
D = \(\sqrt{2+\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}}\)
D = \(\sqrt{2+\sqrt{3}+2-\sqrt{3}}=\sqrt{4}=2\)
E = \(\sqrt{2-\sqrt{2\sqrt{5}-2}}-\sqrt{2+\sqrt{2\sqrt{5}-2}}\)
E2 = \(2-\sqrt{2\sqrt{5}-2}+2+\sqrt{2\sqrt{5}-2}-2\sqrt{\left(2-\sqrt{2\sqrt{5}-2}\right)\left(2+\sqrt{2\sqrt{5}-2}\right)}\)
E2 = \(4-2\sqrt{\left(4-2\sqrt{5}+2\right)}=4-2\sqrt{\left(\sqrt{5}-1\right)}^2=4-2\sqrt{5}+2=\left(\sqrt{5}-1\right)^2\)
=> E = \(\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)