Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
a) \(\dfrac{5}{7}-1\dfrac{4}{7}\left(450\%+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)
\(\dfrac{5}{7}-\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)
\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{5}{7}+\dfrac{1}{14}\)
\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{11}{14}\)
\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}.\dfrac{7}{11}\)
\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{9}{2}=-4\)
\(x=-4:\dfrac{2}{3}=-4.\dfrac{3}{2}=-6\)
Vậy x = \(-6\)
b) \(100=6.7^{\left|x+2\right|}-194\)
\(100+194=6.7^{\left|x+2\right|}\)
\(294=6.7^{\left|x+2\right|}\)
\(294:6=49=7^{\left|x+2\right|}\)
\(\Rightarrow7^2=7^{\left|x+2\right|}\)
\(\Rightarrow2=\left|x+2\right|\Rightarrow\pm2=x+2\)
+ x + 2 = -2 \(\Rightarrow\) x = - 4
+ x + 2 = 2 \(\Rightarrow\) x = 0
Vậy x = - 4 hoặc 0
a)
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)
e)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)
f)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Lời giảiL
$A=1+x+x^2+...+x^n$
$xA=x+x^2+x^3+...+x^n+x^{n+1}$
$\Rightarrow xA-A=(x+x^2+x^3+...+x^{n+1})-(1+x+x^2+...+x^n)$
Hay $A(x-1)=x^{n+1}-1$
$\Rightarrow A=\frac{x^{n+1}-1}{x-1}$ với $x$ nguyên dương khác $1$
Vì $A$ nguyên với mọi $x$ nguyên dương, $n$ tự nhiên nên $\frac{x^{n+1}-1}{x-1}$ nguyên
$\Rightarrow x^{n+1}-1\vdots x-1$ (đpcm)
- Đúng rồi thầy/cô, con đang mong câu trả lời đó đấy ạ :)