Tìm x: (x-1)3=x-1. Các bạn ơi, giúp mình với!
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$x^2+x+1\vdots x+1$
$\Rightarrow x(x+1)+1\vdots x+1$
$\Rightarrow 1\vdots x+1$
$\Rightarrow x+1\in \left\{1; -1\right\}$
$\Rightarrow x\in \left\{0; -2\right\}$
x\(^2\)+x+1⋮x+1
=x(x+1)+1⋮x+1
=1⋮x+1
=x+1∈{1;−1}
=x∈{0;−2}
![](https://rs.olm.vn/images/avt/0.png?1311)
(x + 3) + (x + 7) + (x + 11) + ... + (x + 79) = 860
=> x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
=> (x + x + x + ... + x) + (3 + 7 + 11 + ... + 79) = 860
=> 20x + (79 + 3).20 : 2 = 860
=> 20x + 82.20 : 2 = 860
=> 20x + 82.10 = 860
=> 20x + 820 = 860
=> 20x = 40
=> x = 2
vậy_
#)Giải :
\(\left(x+3\right)+\left(x+7\right)+...+\left(x+79\right)=860\)
\(\left(x+x+...+x\right)+\left(3+7+...+79\right)=860\)(trong mỗi ngoặc có 20 số hạng)
\(x\times20+\frac{\left(79+3\right)\times20}{2}=860\)
\(x\times20+820=860\)
\(x\times20=860-820\)
\(x\times20=40\)
\(x=40\div20\)
\(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right)^3-3\left(x-1\right)=\left(x-1\right)^3-2x\left(1-3x\right)\)
\(x^2+4x+4-3x+3=x^3-3x^2+3x-1-2x+6x^2\)
\(x^2+x+7=x^3+3x^2+x-1\)
\(x^3+3x^2+x-1-x^2-x-7=0\)
\(x^3+2x^2-8=0\)
Đề bài có sai ko bn
![](https://rs.olm.vn/images/avt/0.png?1311)
|x-1|=1
=>x-1 =1 hoặc x-1=-1
TH1: x-1=1 TH2: x-1=-1
=>x=2 =>x=0
vậy x=2,x=0
\(|x-1|=1\)
\(\Rightarrow x-1=1\)
và \(x-1=-1\)
Nếu \(x-1=1\)thì:
\(x=1+1\)
\(x=2\)
Nếu \(x-1=-1\)thì:
\(x=-1+1\)
\(x=0\)
Vậy \(x\in\hept{\begin{cases}2\\0\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: x+ 1+ x+ 2+ 3 = 123
=>2\(\times\) x+6 =123
=> x= (123-6):2
x=117/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : |x + 15| \(\ge0\forall x\)
|8 - y| \(\ge0\forall x\)
Nên C = |x + 15| + |8 - y| \(\ge0\forall x\)
Vậy Cmin là 0 khi x = -15 và y = 8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow x+x+...+x+1+2+...+20=2023\)
\(\Rightarrow10x+20.21:2=2023\Rightarrow10x+210=2023\Rightarrow10x=1813\Rightarrow x=\dfrac{1813}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{1000}\right)\)
`=`\(\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{1000}{1000}-\dfrac{1}{1000}\right)\)
`=`\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{999}{1000}\)
`=`\(\dfrac{1}{1000}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.......\dfrac{99}{100}=\dfrac{1}{100}\)