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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}
\(\left(x+2\right)^3-3\left(x-1\right)=\left(x-1\right)^3-2x\left(1-3x\right)\)
\(x^2+4x+4-3x+3=x^3-3x^2+3x-1-2x+6x^2\)
\(x^2+x+7=x^3+3x^2+x-1\)
\(x^3+3x^2+x-1-x^2-x-7=0\)
\(x^3+2x^2-8=0\)
Đề bài có sai ko bn
|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 \(x\in\hept{\begin{cases}2\\0\end{cases}}\)
\(\Rightarrow x+x+...+x+1+2+...+20=2023\)
\(\Rightarrow10x+20.21:2=2023\Rightarrow10x+210=2023\Rightarrow10x=1813\Rightarrow x=\dfrac{1813}{10}\)
`@` `\text {Ans}`
`\downarrow`
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{1000}\right)\)
`=`\(\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{1000}{1000}-\dfrac{1}{1000}\right)\)
`=`\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{999}{1000}\)
`=`\(\dfrac{1}{1000}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.......\dfrac{99}{100}=\dfrac{1}{100}\)
( 1 + x ) + ( 2 + x ) + ( 3 + x ) + ...+ ( 100 + x ) = 2018
\(\Rightarrow\)( 1 + 2 + 3 + ... + 100 ) + ( x + x + x + ... + x ) = 2018
\(\Rightarrow\){( 1 + 100 ) . [( 100 - 1 ) : 1 + 1 ] : 2 } + ( x + x + ... + x ) = 2018
\(\Rightarrow\)5050 + x . 100 = 2018
\(\Rightarrow\) x100 = 2018 - 5050 = -3032
\(\Rightarrow\)x = -3032 : 100 = -30,32
vậy x = -30,32
mà nè sai thì xin lỗi đề hơi có vấn đề nếu sai thì sorry nha !!!
-12(x-5)+7(3-x)=5
=> -12x+60+21-7x=5
=> -19x+81=5
=>-19x=5-81
=>-19x=-76
=>x=4
Vậy x=4
3/2.(x-5/3)+4/5=x+1
3/2.x-3/2.5/3+4/5=x+1
3/2.x-5/2-x=1+4/5
3/2.x-5/2-x=9/5
3/2.x-x=9/5+5/2
1/2.x=43/10
x=43/10:1/2
x=43/5