


a)
3
n...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(\frac{1}{2^2}< \frac{1}{1.2}\) \(\frac{1}{3^2}< \frac{1}{2.3}\) ........ \(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\) => \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\) Đpcm a) |2x + 1| - 19 = -7 => \(\left|2x+1\right|=-7+19=12\) => \(\left[{}\begin{matrix}2x+1=12\\2x+1=-12\end{matrix}\right.\) => \(\left[{}\begin{matrix}2x=12-1=11\\2x=-12-1=-13\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{11}{2}\\x=-\frac{13}{2}\end{matrix}\right.\) Vậy:............ b) -28 – 7. |- 3x + 15| = -70 => \(\text{7. |- 3x + 15| = -28 - (-70) = -28 + 70 = 42}\) => \(\left|-3x+15\right|=42:7=6\) => \(\left[{}\begin{matrix}-3x+15=6\\-3x+15=-6\end{matrix}\right.\) => \(\left[{}\begin{matrix}-3x=6-15=-9\\-3x=-6-15=-6+\left(-15\right)=-21\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-9:\left(-3\right)=3\\x=x=-21:\left(-3\right)=7\end{matrix}\right.\) Vậy:..................... c) |18 – 2. |-x + 5|| = 12 => \(\left[{}\begin{matrix}18-2.\left|-x+5\right|=12\\18-2.\left|-x+5\right|=-12\end{matrix}\right.\) => \(\left[{}\begin{matrix}2.\left|-x+5\right|=18-12=6\\2.\left|-x+5\right|=18-\left(-12\right)=18+12=30\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left|-x+5\right|=6:2=3\\\left|-x+5\right|=30:2=15\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x+5=3\\-x+5=-3\end{matrix}\right.\\\left[{}\begin{matrix}-x+5=15\\-x+5=-15\end{matrix}\right.\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x=3-5=-2\\-x=-3-5=-8\end{matrix}\right.\\\left[{}\begin{matrix}-x=15-5=10\\-x=-15-5=-20\end{matrix}\right.\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\\\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\end{matrix}\right.\) n.(2x-5)2=9 (2x-5)2=32 * 2x-5=3 * 2x-5=-3 2x=3+5 2x=-3+5 2x=8 2x=2 x=8:2 x=2:2 x=4 x=1 vậy x=4 hoặc x=1 o.(1-3x )3=-8 (1-3x)3=(-2)3 1-3x=-2 3x=1-(-2) 3x=3 x=3:3 x=1 vậy x=1 So sánh : a ) 31^11 và 17^14 31^11 < 32^11= (25)11 = 2^55 => 31^11 < 2^55 17^14>16^14=(24)14 = 2^56 =>17^14>2^56 =>31^11 < 2^55 < 2^56 < 17^14 =>31^11 < 17^14 b ) 3^500 và 7^300 3^500 = ( 35)100 = 243100 7^300 = ( 73)100 = 343100 => 243100 < 343100 => 3^500 < 7^300 Tìm x : a ) 2x . 4 = 128 => 2x = 32 => 2x = 25 => x = 5 b ) 2x . 22 = ( 23)2 = 64 => 2x = 64 : 22 = 16 => 2x = 24 => x = 4 Bài cuối bạn tham khảo tại : Câu hỏi của Linh Phan - Toán lớp 6 - Học toán với OnlineMath Link : https://olm.vn/hoi-dap/detail/198524999512.html Bài 1: \(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\) \(=x^{1+2+3+4+5+...+49+50}\) \(=x^{\frac{51.50}{2}}\) \(=x^{1275}\) \(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\) \(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\) \(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\) Bài 2: Tìm x \(\left(x-1\right)^4:3^2=3^6\) \(\Rightarrow\left(x-1\right)^4=3^6\times3^2\) \(\Rightarrow\left(x-1\right)^4=3^8\) \(\Rightarrow\left(x-1\right)^4=3^{2.4}\) \(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\) \(\Rightarrow x-1=9\) \(\Rightarrow x=10\) Bài 3 và bài 4 mk làm sau Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính) b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi Bài 2 : \(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\) Bài 3 : \(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt



\(\text{b) Ta có:}\)