Tìm x, biết : 2x + 7 \(⋮\)x + 1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
20 tháng 9 2021
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
30 tháng 9 2016
\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{7^x.\left(7^2+7+1\right)}{57}=7^x\)
\(\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}=\frac{5^{2x}\left(1+5+5^3\right)}{131}=\frac{25^x.131}{131}=25^x\)
\(\Rightarrow7^x=25^x\Rightarrow x=0\)
QT
1
12 tháng 12 2023
\(\left(x+7\right)+3x\left(2x-1\right)-2x\left(3x+15\right)=-42\)
=>\(x+7+6x^2-3x-6x^2-30x=-42\)
=>\(-32x=-42-7=-49\)
=>\(x=\dfrac{49}{32}\)
12 tháng 3 2022
a) \(\left(2x-1\right)^3=27\)
\(\Rightarrow\left(2x-1\right)^3=3^3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=3+1=4\)
\(\Rightarrow x=4:2=2\)
b) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1=4\)
\(\Rightarrow x=4:2=2\)
c) \(\left(x+2\right)^3=\left(2x\right)^3\)
\(\Rightarrow x+2=2x\)
\(\Rightarrow2=2x-2=1x\)
\(\Rightarrow x=2:1=2\)
d) \(\left(2x-1\right)^7=x^7\)
\(\Rightarrow2x-1=x\)
\(\Rightarrow-1=x-2x\)
\(\Rightarrow-1=-x\)
\(\Rightarrow x=1\)
\(\frac{2x+7}{x+1}\)
\(=\frac{2x+2+5}{x+1}\)
\(=\frac{2x+2}{x+1}+\frac{5}{x+1}\)
\(=2+\frac{5}{x+1}\)
Để 2x + 7 chia hết cho x + 1 thì 5 phải chia hết cho x+ 1
\(\Rightarrow x+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(x+1=-5\Rightarrow x=-6\)
\(x+1=-1\Rightarrow x=-2\)
\(x+1=1\Rightarrow x=0\)
\(x+1=5\Rightarrow x=4\)
Vậy x = -6 ; -2 ; 0 ; 4
2x + 7 \(⋮\) x+1
Ta có : 2x+7 = 2(x+1) + 5
mà 2(x+1) \(⋮\) x+1 để 2x + 7 \(⋮\) x+1
=> 5\(⋮\) x+1 hay x+1 \(\in\) Ư(5) = {1;-1;5;-5}
Ta có bảng sau
Vây x\(\in\) {0;-2;4;-6}