Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(0,49\right)^{36}=\left[\left(0,7\right)^2\right]^{36}=\left(0,7\right)^{72}\)
\(\left(0,512\right)^{24}=\left[\left(0,8\right)^3\right]^{24}=\left(0,8\right)^{72}\)
Vì 0,7<0,8 suy ra \(\left(0,7\right)^{72}< \left(0,8\right)^{72}\)
Vậy \(\left(0,49\right)^{36}< \left(0,512\right)^{24}\)
(0,49)36=[( 0,7)2]36=(0,7)72 (1)
(0,512)24=[(0,8)3]24=(0,8)72(20
Từ (1) và (2) vậy (0,49)36<(0,512)24
Ta có:
\(\sqrt{0,49}+\sqrt{\frac{25}{36}}=\sqrt{0,7^2}+\sqrt{\left(\frac{5}{6}\right)^2}=0,7+\frac{5}{6}=\frac{23}{15}\)
Bạn Duy giúp mik vs
Tính:
a) \(\sqrt{0,36}+\sqrt{0,49}\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}\)
a)\(\sqrt{0,36}\)+\(\sqrt{0,49}\)=0,6+0,7=1,3
b)\(\sqrt{\frac{4}{9}}\)-\(\sqrt{\frac{25}{36}}\)=2/3-5/6=4/6-5/6=-1/6
a) \(\sqrt{0,36}+\sqrt{0,49}=\sqrt{\left(0,6\right)^2}+\sqrt{\left(0,7\right)^2}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\sqrt{\left(\frac{2}{3}\right)^2}-\sqrt{\left(\frac{5}{6}\right)^2}=\frac{2}{3}-\frac{5}{6}=-\frac{1}{6}\)
536 = (53)12 = 12512
1124 = (112)12 = 12112
12512 > 12112 nên 536 > 1124
\(\left(\dfrac{1}{2}\right)^{24}=\left[\left(\dfrac{1}{2}\right)^2\right]^{12}=\left(\dfrac{1}{4}\right)^{12}\\ \left(\dfrac{1}{3}\right)^{36}=\left[\left(\dfrac{1}{3}\right)^3\right]^{12}=\left(\dfrac{1}{27}\right)^{12}\\ Vì:\dfrac{1}{4}>\dfrac{1}{27}\Rightarrow\left(\dfrac{1}{4}\right)^{12}>\left(\dfrac{1}{27}\right)^{12}\Rightarrow\left(\dfrac{1}{2}\right)^{24}>\left(\dfrac{1}{3}\right)^{36}\)
Dễ thấy \(2^{24}< 3^{36}\Rightarrow\dfrac{1}{2^{24}}>\dfrac{1}{3^{36}}\)
\(\dfrac{12}{18}=\dfrac{24}{36}=\dfrac{72}{108}=\dfrac{12+24+72}{18+36+108}=\dfrac{12-24+72}{18-36+108}\)
a)
(0,49)36=[(0,49)3]12=(0,117649)12
(0,512)24=[(0,512)2]12=(0,262144)12
=>(0,49)36<(0,512)24