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KK
5
WR
30 tháng 6 2019
đề là \(x^2-\frac{1}{x^2}\)hay là \(x^2+\frac{1}{x^2}\)vậy? Xem lại đề thử xem!
30 tháng 6 2019
\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=4\)
\(\Leftrightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\left(x;y\right)=\left(1;1\right);\left(1;-1\right);\left(-1;1\right);\left(-1;-1\right)\)
19 tháng 7 2019
bài 2
làm câu B;C nha
B)
\(27^3=\left(3^3\right)^3=3^9\)
\(9^5=\left(3^2\right)^5=3^{10}\)
vì \(10>9\)
\(=>9^5>27^3\)
C)
\(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2^3}\right)^6=\frac{1^6}{2^{18}}=\frac{1}{2^{18}}\)
\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2^5}\right)^4=\frac{1^4}{2^{20}}=\frac{1}{2^{20}}\)
vì \(2^{18}< 2^{20}\)
\(=>\frac{1}{2^{18}}>\frac{1}{2^{20}}\)
\(=>\left(\frac{1}{8}\right)^6>\left(\frac{1}{32}\right)^4\)
19 tháng 7 2019
\(\text{A.}\frac{32^3.9^5}{8^3.6^6}=\frac{\left(2^5\right)^3.\left(3^2\right)^5}{\left(2^3\right)^3.\left(2.3\right)^6}=\frac{2^{15}.3^{10}}{2^9.2^6.3^6}=\frac{3^{10}}{3^6}=3^4=81\)
\(\text{B.}\frac{\left(5^5-5^4\right)^3}{50^6}=\frac{2500^3}{50^6}=\frac{\left(50^2\right)^3}{50^6}=\frac{50^6}{50^6}=1\)
Bài 2:
\(\text{A.Ta có:}\)
\(5^6=\left(5^3\right)^2=125^2\)
\(\left(-2\right)^{14}=2^{14}=\left(2^7\right)^2=128^2\)
Vì \(125< 128\)
\(\Rightarrow125^2< 128^2\)
\(\Rightarrow5^6< \left(-2\right)^{14}\)
\(\text{B.Ta có:}\)
\(9^5=\left(3^2\right)^5=3^{10}\)
\(27^3=\left(3^3\right)^3=3^9\)
Vì \(9< 10\)
\(\Rightarrow3^9< 3^{10}\)
\(\Rightarrow27^3< 9^5\)
\(\text{C.Ta có:}\)
\(\left(\frac{1}{8}\right)^6=\left[\left(\frac{1}{2}\right)^3\right]^6=\left(\frac{1}{2}\right)^{18}\)
\(\left(\frac{1}{32}\right)^4=\left[\left(\frac{1}{2}\right)^5\right]^4=\left(\frac{1}{2}\right)^{20}\)
Vì \(18< 20\)
\(\Rightarrow\left(\frac{1}{2}\right)^{18}< \left(\frac{1}{2}\right)^{20}\)
\(\Rightarrow\left(\frac{1}{8}\right)^6< \left(\frac{1}{32}\right)^4\)
SA
25 tháng 10 2018
\(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}.\left[\left(-3\right)^2\right]^{25}-\dfrac{2}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{1}{3}\right)^{50}.\left(-3\right)^{50}-\dfrac{2}{12}\)
\(=\left[\dfrac{1}{3}.\left(-3\right)\right]^{50}-\dfrac{1}{6}\)
\(=\left(-1\right)^{50}-\dfrac{1}{6}\)
\(=1-\dfrac{1}{6}\)
\(=\dfrac{6}{6}-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
25 tháng 10 2018
\(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4=\)
=\(\dfrac{1}{3^{50}}\cdot\left(-9\right)^{25}-\dfrac{2}{3}\cdot\dfrac{1}{4}\)
=\(\dfrac{\left(-9\right)^{25}}{9^{25}}-\dfrac{1}{6}\)
=\(-1-\dfrac{1}{6}\)
=\(-\dfrac{6}{6}-\dfrac{1}{6}\)
=\(-\dfrac{7}{6}\)
DH
4 tháng 9 2017
a, \(\left(m-2\right)^2=1\)
\(\Rightarrow m-2\in\left\{-1;1\right\}\Rightarrow m\in\left\{1;3\right\}\)
Vậy..............
b, \(\left(2m-1\right)^3=-8\)
\(\Rightarrow2m-1=-2\Rightarrow2m=-1\Rightarrow m=-\dfrac{1}{2}\)
Vậy.................
c, \(\left(m+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow m+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow m\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
Vậy.................
Chúc bạn học tốt!!!
NT
0
27 tháng 12 2016
a) \(M=\frac{-2.10.x^{1+2}y^{2+2}}{5}=-4.x^3.y^4\) Bậc 4 y bậc 3 với x
b) \(N=4.\frac{\left(-1\right)^2}{2^2}x^{3+2}.y^{1+4}=x^5.y^5\) bậc 5 cả x, y
25 tháng 9 2020
a) Ta có: \(\left(8+\frac{1}{2}\right)^3:\left(1+\frac{1}{2}\right)^2\)
\(=\left(\frac{17}{2}\right)^3:\left(\frac{3}{2}\right)^2\)
\(=\frac{17^3}{8}\cdot\frac{2^2}{3}\)
\(=\frac{17^3}{2\cdot3}=\frac{4913}{6}\)
b) Ta có: \(\left(4^4-8^3\right):2^7\)
\(=\frac{2^8-2^9}{2^7}\)
\(=\frac{2^8\left(1-2\right)}{2^7}\)
\(=-2\)
, bạn nào làm đc mik tick cho
18 tháng 7 2017
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
18 tháng 7 2017
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi
=\(\frac{1}{64}\)