Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)

Ta có: \(\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2019}-1+11}{10^{2020}-1+11}=\frac{10^{2019}+10}{10^{2020}+10}=\frac{10.\left(10^{2018}+1\right)}{10.\left(10^{2019}+1\right)}=\frac{10^{2018}+1}{10^{2019}+1}\)

\(\Rightarrow\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2018}+1}{10^{2019}+1}\)

Đặt \(A=\frac{10^{2019}-1}{10^{2020}-1}\)

\(B=\frac{10^{2018}+1}{10^{2019}+1}\)

Dễ thấy \(A< 1\)

Áp dụng kết quả bài trên nếu \(\frac{a}{b}< 1\)thì \(\frac{a+m}{b+m}>\frac{a}{b}\)với m>0

Vậy \(A=\frac{10^{2019}-1}{10^{2020}-1}< \frac{\left[10^{2019}-1\right]+11}{\left[10^{2020}-1\right]+11}=\frac{10^{2019}+10}{10^{2020}+10}\)

\(A< \frac{10\left[10^{2018}+1\right]}{10\left[10^{2019}+1\right]}=\frac{10^{2018}+1}{10^{2019}+1}=B\)

Do đó : A<B

\(x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

Vậy \(x\in\left\{0;-3\right\}\)

\(x.\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-3\end{cases}}}\)

vậy \(x=0\)hoặc\(x=-3\)

a) \(5^{30}=\left(5^3\right)^{10}=125^{10}\)

Vì \(125^{10}>124^{10}\Rightarrow5^{30}>124^{10}\)

b) \(3^2+4^2=9+16=25\)

\(\left(3+4\right)^2=7^2=49\)

Mà \(25< 49\Rightarrow3^2+4^2< \left(3+4\right)^2\)

\(x^2-3xy+3y-x=\)\(1\)

\(\Leftrightarrow\left(x^2-x\right)+\left(-3xy+3y\right)=1\)

\(\Leftrightarrow x\left(x-1\right)-3y\left(x-1\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(x-3y\right)=1\)

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

+)  \(\hept{\begin{cases}x-1=-1\\x-3y=-1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x-3y=-1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\y=\frac{1}{3}\end{cases}}\)

Vậy (x;y)= (2;1/3 ) ,(0; 1/3)

\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x-1=1\\x-3y=1\end{cases}}\\\hept{\begin{cases}x-1=-1\\x-3y=-1\end{cases}}\end{cases}}\)

2 . 5³ - 36 - 3²

= 2 . 125 - 36 - 9

= 250 - 36 - 9

= 214 - 9

= 205

b) 13 . 3 . 5² + 13 - 25 - 140

= 13 . 3 . 25 + 13 - 25 - 140

= 39 . 25 + 13 - 25 - 140

= 975 + 13 - 25 - 140

= 823

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)

= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

= \(\frac{1}{1.2}-\frac{1}{49.50}\)

= \(\frac{1}{2}-\frac{1}{2450}\)

= \(\frac{612}{1225}\)

đặt

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\)

\(\Rightarrow\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{1.2}-\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2450}=\frac{621}{1225}\)

\(\Rightarrow A=\frac{306}{1225}\)

a) \(5^2.21-5^2.16\)

= \(5^2.\left(21-16\right)\)

= \(5^2.5\)

= \(5^3\)

a,  (7^2)^3:7^4=7^6:7^4=7^2

b,   (2^2)^5:2^7=2^10:2^7=2^3

Trả lời

a)49³:7⁴=7⁶:7⁴=7²=49

b)4⁵:2⁷=2¹⁰:2⁷=2³=8

Hok tốt !

Giúp em với các CTV 

a)  Ta có : \(n^2⋮n-3\)

\(\Rightarrow n^2-3^2+3^2⋮n-3\)

\(\Rightarrow\left(n^2-3^2\right)+3^2⋮n-3\)

\(\Rightarrow\left(n-3\right)\left(n+3\right)+3^2⋮n-3\)(sử dụng hằng đẳng thức trừ 2 bình phương của 2 số)

Vì \(\left(n-3\right)\left(n+3\right)⋮n-3\)

\(\Rightarrow3^2⋮n-3\)

\(\Rightarrow9⋮n-3\)

\(\Rightarrow n-3\inƯ\left(9\right)\)

\(\Rightarrow n-3\in\left\{\pm1;\pm3;\pm9\right\}\)

Lập bảng xét các trường hợp : 

Vậy các \(n\inℕ\)thỏa mãn là : 4;2;6;0;12