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NT
2
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 ...
TV
http://lazi.vn/uploads/edu/exercise/1510324551_13.jpg
0
8 tháng 7 2021
\(-7xy\sqrt{\frac{16}{xy}}\)
\(-7xy\frac{4\sqrt{xy}}{xy}\)
\(-28\sqrt{xy}\)
LH
![[âIMG]](https://lazi.vn/uploads/edu/exercise/1505311489_8.jpg)
3
HN
1
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 \(B=3\)
NT
giúp mình với nha - Mình cảm ơn!!
2
TT
2
\(1,\sqrt{2}\left(3-2\sqrt{9}+2\sqrt{16}-\sqrt{25}\right)=\sqrt{2}\left(3-6+8-5\right)\)
\(=\sqrt{2}.0=0\)
\(2,\sqrt{2}\left(\sqrt{25}-\sqrt{9}+\sqrt{100}-\sqrt{81}\right)\)
\(=\sqrt{2}\left(5-3+10-9\right)=3\sqrt{2}\)
\(3,\sqrt{5}\left(5+\sqrt{4}-3\sqrt{9}\right)=\sqrt{5}\left(5+2-9\right)=-2\sqrt{5}\)
\(4,\sqrt{3}\left(5\sqrt{16}-4\sqrt{9}-2\sqrt{25}+\sqrt{36}\right)\)
\(=\sqrt{3}\left(20-12-10+6\right)=4\sqrt{3}\)
\(5,\sqrt{12}-\sqrt{300}-\sqrt{3}+\frac{10\sqrt{3}}{3}\)
\(\sqrt{3}\left(\sqrt{4}-\sqrt{100}-1+\frac{10}{3}\right)=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)=-\frac{17\sqrt{3}}{3}\)
\(6,\sqrt{3}\left(3\sqrt{4}-4\sqrt{9}+5\sqrt{16}\right)=\sqrt{3}\left(6-12+20\right)\)
\(=14\sqrt{3}\)
\(7,\sqrt{3}\left(2+5-4\right)=3\sqrt{3}\)
\(8,\sqrt{2}\left(8+8-15\right)=\sqrt{2}\)
\(9,\sqrt{5}\left(6-6+4\right)=4\sqrt{5}\)
\(10,\sqrt{6}\left(3-6+3-5\right)=-5\sqrt{6}\)
11, \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
\(=2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
\(=6\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
\(=8\sqrt{2}\)
12, \(3\sqrt{8}-4\sqrt{18}+5\sqrt{32}-\sqrt{50}\)
\(=3.2\sqrt{2}-4.3\sqrt{2}+5.4\sqrt{2}-5\sqrt{2}\)
\(=6\sqrt{2}-12\sqrt{2}+20\sqrt{2}-5\sqrt{2}\)
\(=9\sqrt{2}\)
13, \(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
\(=5\sqrt{5}-2.2\sqrt{5}-3.4\sqrt{5}+4.3\sqrt{5}\)
\(=5\sqrt{5}-4\sqrt{5}-12\sqrt{5}+12\sqrt{5}\)
\(=\sqrt{5}\)
14, \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}\)
\(=2.2\sqrt{7}+2.3\sqrt{7}-3.5\sqrt{7}+4\sqrt{7}\)
\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}\)
\(=-\sqrt{7}\)
15, \(3\sqrt{2}+\sqrt{8}+\frac{1}{2}\sqrt{50}-\sqrt{32}\)
\(=3\sqrt{2}+2\sqrt{2}+\frac{1}{2}.5\sqrt{2}-4\sqrt{2}\)
\(=3\sqrt{2}+2\sqrt{2}+\frac{5}{2}\sqrt{2}-4\sqrt{2}\)
\(=\frac{7}{2}\sqrt{2}\)
16, \(3\sqrt{50}-2\sqrt{12}-\sqrt{18}+\sqrt{75}-\sqrt{8}\)
\(=3.5\sqrt{2}-2.2\sqrt{3}-3\sqrt{2}+5\sqrt{3}-2\sqrt{2}\)
\(=15\sqrt{2}-4\sqrt{3}-3\sqrt{2}+5\sqrt{3}-2\sqrt{2}\)
\(=10\sqrt{2}+\sqrt{3}\)
17, \(2\sqrt{75}-3\sqrt{12}+\sqrt{27}\)
\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)
\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)
\(=7\sqrt{3}\)
18, \(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
\(=2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
\(=4\sqrt{3}\)
19, \(\sqrt{27}-\sqrt{12}+\sqrt{75}+\sqrt{147}\)
\(=3\sqrt{3}-2\sqrt{3}+5\sqrt{3}+7\sqrt{3}\)
\(=13\sqrt{3}\)
20, \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
\(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}\)
\(=-8\sqrt{3}\)
21, \(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\)
\(=6.\frac{2\sqrt{2}}{3}-5.\frac{4\sqrt{2}}{5}+14.\frac{3\sqrt{2}}{7}\)
\(=4\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)
\(=6\sqrt{2}\)