K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 11 2017

Cách 1: 3√1728 : 3√64 = 12:4 = 3

Cách 2: 3√1728:3√64 = 3√(1728/64) = 3√27 = 3

11 tháng 7 2016

a) \(A=\sqrt{64}+4\sqrt{4}+2016=\sqrt{8^2}+4.\sqrt{2^2}+2016=8+4.2+2016=2032\)

b) \(B=2\sqrt{8}-3\sqrt{18}+4\sqrt{128}-5\sqrt{32}=4\sqrt{2}-9\sqrt{2}+32\sqrt{2}-20\sqrt{2}\)

\(=\sqrt{2}\left(4-9+32-20\right)=7\sqrt{2}\)

a,

\(A=\sqrt{8}^2+2.\sqrt{8}.\sqrt{2}+\sqrt{2}^2+2014\)

\(=\left(\sqrt{8}+\sqrt{2}\right)^2+2014\)

9 tháng 8 2016

  goi V la` can bac hai , abs la` gia tri tuyet doi 
ta co P=V((x^3+3)^2/x^2) + V(x-2)^2 =abs((x^3+3)/x)+abs(x-2) 
do x thuoc Z nen abs(x-2) thuoc Z 
vay de~ P thuoc Z thi` (x^3+3) chia het cho x 
=>x thuoc uoc cua 3 
=>X={-3;-1;1;3} =>S={5;11;13}

16 tháng 9 2018

\(a.\sqrt{0,4}.\sqrt{64}=\sqrt{0,4.64}=\sqrt{25,6}\)

\(b.\sqrt{5,2}.\sqrt{1,3}=\sqrt{5,2.1,3}=\sqrt{6,76}=2.6\)

\(c.\sqrt{12,1}.\sqrt{360}=\sqrt{12,1.360}=\sqrt{4356}=66\)

a: \(\Leftrightarrow2x+3=14-6\sqrt{5}\)

=>2x=11-6 căn 5

hay \(x=\dfrac{11-6\sqrt{5}}{2}\)

b: \(\Leftrightarrow\sqrt{7x}+5=11+4\sqrt{7}\)

=>căn 7x=6+4 căn 7

=>\(x=\dfrac{\left(6+4\sqrt{7}\right)^2}{7}\)

d: \(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

=>-căn x-1=-17

=>căn x-1=17

=>x-1=289

=>x=290

18 tháng 10 2020

a) \(\sqrt{36}.\sqrt{121}+\sqrt[3]{-64}-\sqrt[3]{125}\)

\(=6.11+\left(-4\right)-5=66-9=57\)

b) \(\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}-30\sqrt{\frac{3}{25}}\)

\(=\sqrt{25.3}+\left|\sqrt{3}-2\right|-30.\frac{\sqrt{3}}{\sqrt{25}}\)

\(=5\sqrt{3}+2-\sqrt{3}-30.\frac{\sqrt{3}}{5}\)

\(=5\sqrt{3}+2-\sqrt{3}-6\sqrt{3}=2-2\sqrt{3}\)

c) \(\sqrt{11-4\sqrt{7}}-\frac{12}{1+\sqrt{7}}=\sqrt{7-4\sqrt{7}+4}-\frac{12}{1+\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\frac{12}{1+\sqrt{7}}=\left|\sqrt{7}-2\right|-\frac{12}{1+\sqrt{7}}\)

\(=\left(\sqrt{7}-2\right)-\frac{12}{\sqrt{7}+1}=\frac{\left(\sqrt{7}-2\right)\left(\sqrt{7}+1\right)}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}\)

\(=\frac{5-\sqrt{7}}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}=\frac{-7-\sqrt{7}}{\sqrt{7}+1}\)

\(=\frac{-\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}=-\sqrt{7}\)

10 tháng 10 2020

a, \(=2\sqrt{7}-8+15\sqrt{7}-12=17\sqrt{7}-20\)

b, \(=2\sqrt{2}-10\sqrt{2}+4\sqrt{2}=-4\sqrt{2}\)

c, \(=\frac{3}{8}.\frac{4}{3}-2.\frac{2}{5}=\frac{1}{2}-\frac{4}{5}=-\frac{3}{10}\)

d, \(\sqrt{\left(\sqrt{3-1}\right)^2}-\sqrt{\left(\sqrt{3-2}\right)^2}=\sqrt{3-1}-\sqrt{3-2}=\sqrt{2}-\sqrt{1}=\sqrt{2}-1\)

e, \(\sqrt{2-3}\) không tồn tại

14 tháng 7 2016

\(A=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}=\sqrt{3^2-\left(\sqrt{5}\right)^2}=\sqrt{4}=2\)

\(B=\sqrt{150.27.96}=\sqrt{150}.\sqrt{27}.\sqrt{96}=5\sqrt{6}.3\sqrt{3}.4\sqrt{6}=360\sqrt{3}\)

\(C=\left(\sqrt{27}+\sqrt{48}\right)^2-\left(\sqrt{27}-\sqrt{48}\right)^2\)\(=\left[\left(\sqrt{27}+\sqrt{48}-\sqrt{27}+\sqrt{48}\right)\left(\sqrt{27}+\sqrt{48}+\sqrt{27}-\sqrt{48}\right)\right]\)

\(=2\sqrt{27}.2\sqrt{48}=2.3\sqrt{3}.2.4\sqrt{3}=144\)

\(D=\sqrt{137^2-88^2}-\sqrt{192^2-111^2}=\sqrt{\left(137+88\right)\left(137-88\right)}-\sqrt{\left(192+111\right)\left(192-111\right)}\)

\(=\sqrt{225.49}-\sqrt{303.81}=15.7-9.\sqrt{303}=9\left(\frac{35}{3}-\sqrt{303}\right)\)

14 tháng 7 2016

\(E=\sqrt{\frac{225}{4}.\frac{81}{25}.\frac{49}{64}}=\frac{15}{2}.\frac{9}{5}.\frac{7}{8}=\frac{189}{16}\)

\(F=\sqrt{\frac{27}{25}}.\sqrt{\frac{49}{189}}.\sqrt{\frac{700}{99}}=\frac{3\sqrt{3}}{5}.\frac{7}{3\sqrt{21`}}.\frac{10\sqrt{7}}{3\sqrt{11}}=\frac{14}{3\sqrt{11}}\)

\(H=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{\frac{35}{5}}+\sqrt{\frac{21}{5}}\right]=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{7}+\sqrt{\frac{21}{5}}\right]\)

\(=\sqrt{105}.\left[\frac{\sqrt{75}-\sqrt{49}+\sqrt{147}}{\sqrt{35}}\right]=\sqrt{3}\left(12\sqrt{3}-7\right)=36-7\sqrt{3}\)

\(K=\sqrt{64.14.21.54}-\sqrt{35.45.12}=8.\sqrt{14}.\sqrt{21}.3\sqrt{6}-\sqrt{35}.3\sqrt{5}.2\sqrt{3}\)

\(=24.\sqrt{14.21.6}-6\sqrt{35.5.3}=24.42-30\sqrt{21}=30\left(\frac{168}{5}-\sqrt{21}\right)\)