4 8/3 và 3\(\sqrt2\)

4\(\frac83\) = \(\frac{20}{3}\) = \(\sqrt{\frac{400}{3}}\)

3\(\sqrt2\) = \(\sqrt{9.2}\) = \(\sqrt{18}\) = \(\sqrt{\frac{54}{3}}\)

Vì 400/3 > 54/3 nên \(\sqrt{\frac{400}{3}}\) > \(\sqrt{\frac{54}{3}}\)

Vậy: 4\(\frac83\) > 3\(\sqrt2\)

So sánh:

5\(\sqrt{\left(-10\right)^2}\) và 10\(\sqrt{\left(-5\right)^2}\)

5\(\sqrt{\left(-10\right)^2}\) = \(\sqrt{100.5^2}\) = \(\sqrt{2500}\)

10\(\sqrt{\left(-5\right)^2}\) = \(\sqrt{10^2.25}\) = \(\sqrt{2500}\)

Vậy 5\(\sqrt{\left(-10\right)^2}\) = 10\(\sqrt{\left(-5\right)^2}\)

\(\left(\frac{-2}{3}\right)^3=\frac{-8}{27}\)

\(\frac{\left(-2\right)^3}{3^3}=\frac{-8}{27}\)

\(=>\left(-\frac{2}{3}\right)^3=\frac{\left(-2\right)^3}{3^3}\)

Áp dụng câu trên ta được :

 \(\frac{10^5}{2^5}=\left(\frac{10}{2}\right)^5\)

Ủng hộ nha

o0o I am a studious person o0o:cái này có công thức trong sách giáo khoa rồi cần j phải tính

a) 

\(\left(-\frac{2}{3}\right)^3=\frac{\left(-2\right)^3}{3^3}\)

mà \(\frac{\left(-2\right)^3}{3^3}\) là vế phải 

\(\Rightarrow\) \(\left(-\frac{2}{3}\right)^3=\frac{\left(-2\right)^3}{3^3}\) 

b) 

\(\frac{10^5}{2^5}=\left(\frac{10}{2}\right)^5\)

mà \(\left(\frac{10}{2}\right)^5\) là vế phải 

Nên \(\frac{10^5}{2^5}=\left(\frac{10}{2}\right)^5\)

a) \(\left(-\frac{2}{3}\right)^3=\left(-\frac{2}{3}\right).\left(-\frac{2}{3}\right).\left(-\frac{2}{3}\right)=\frac{\left(-2\right).\left(-2\right).\left(-2\right)}{3.3.3}=\frac{\left(-2\right)^3}{3^3}=\frac{-8}{27}\)

b) \(\left(\frac{10}{2}\right)^5=\frac{10}{2}.\frac{10}{2}.\frac{10}{2}.\frac{10}{2}.\frac{10}{2}=\frac{10.10.10.10.10}{2.2.2.2.2}=\frac{10^5}{2^5}=\frac{100000}{32}=3125\)

1) Ta có: \(\left|9y-1\right|+\left(2x+3\right)^2=0\)

Mà \(\hept{\begin{cases}\left|9y-1\right|\ge0\\\left(2x+3\right)^2\ge0\end{cases}}\left(\forall x,y\right)\)

=> \(\left|9y-1\right|+\left(2x+3\right)^2\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|9y-1\right|=0\\\left(2x+3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}9y-1=0\\2x+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)

2)

a) Ta có: \(\left[\left(-\frac{1}{3}\right)^7\right]^4=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)

và \(\left[\left(-\frac{1}{2}\right)^{14}\right]^2=\left(\frac{1}{2}\right)^{28}=\frac{1}{2^{28}}\)

Vì \(\frac{1}{3^{28}}< \frac{1}{2^{28}}\Rightarrow\left[\left(-\frac{1}{3}\right)^7\right]^4< \left[\left(-\frac{1}{2}\right)^{14}\right]^2\)

b) Ta có: \(\left(-\frac{2}{3}\right)^{12}=\left[\left(-\frac{2}{3}\right)^2\right]^6=\left(\frac{4}{9}\right)^6\)

Ta thấy \(0< \frac{4}{9}< 1\)\(\Rightarrow\left(\frac{4}{9}\right)^6>\left(\frac{4}{9}\right)^7\)

\(\Rightarrow\left(-\frac{2}{3}\right)^{12}>\left(\frac{4}{9}\right)^7\)

3. \(\left(\frac{1}{2^5}\right)^{25}=\left(\frac{1^5}{2^5}\right)^{25}=\left[\left(\frac{1}{2}\right)^5\right]^{25}=\left(\frac{1}{2}\right)^{125}\)

\(\left(\frac{1}{3^{25}}\right)^5=\left(\frac{1^{25}}{3^{25}}\right)^5=\left[\left(\frac{1}{3}\right)^{25}\right]^5=\left(\frac{1}{3}\right)^{125}\)

Vì \(\frac{1}{2}>\frac{1}{3}\Rightarrow\left(\frac{1}{2^5}\right)^{25}>\left(\frac{1}{3^{25}}\right)^5\)

1. \(3^{800}=\left(3^8\right)^{100}=6561^{100}\)

\(5^{500}=\left(5^5\right)^{100}=3125^{100}\)

Vì \(6561>3125\Rightarrow3^{800}>5^{500}\)

2. \(\left(-2\right)^{3000}=\left[\left(-2\right)^3\right]^{1000}=\left(-8\right)^{1000}\)

\(\left(-3\right)^{2000}=\left[\left(-3\right)^2\right]^{1000}=9^{1000}\)

Vì \(-8< 9\Rightarrow\left(-2\right)^{3000}< \left(-3\right)^{2000}\)

a) $2,\left(15\right)$ > $2,\left(14\right)$

b) $-0,2673$ > $-0,267\left(3\right)$

c) $1,\left(2357\right)$ > $1,2357$

d) $0,\left(428571\right)$ = $\frac{3}{7}$

a)2,(15)>2,(14)

b)-0,2673>-0,267(3)

c)1,(2357)>1,2357

d)0,(428571)=3/7

Ta có: \(2^{2^3}=2^{\left(2\right)^3}=2^8\)

           \(\left(2^2\right)^3=2^{2.3}=2^6\)

Vì 2⁶<2⁸ => \(2^{2^3}>\left(2^2\right)^3\)