Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=2\)
\(\le\left|x\right|-\left|2\right|+\left|x\right|-\left|3\right|+\left|x\right|-\left|4\right|\)
\(x-2+x-3+x-4=2\)
\(\Leftrightarrow x+x+x-2-3-4=2\)
\(\Leftrightarrow x^3-2-3-4=2\)
\(x^3=2+4+3+2\)
\(x^3=11\)
\(x=\sqrt[3]{11}\)
Có: \(\frac{1}{x\left(x+1\right)}\)= \(\frac{1}{x}-\frac{1}{x+1}\)
Mà \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2017}\)
=> \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2017}\)
=> \(-\frac{1}{x+1}\)= \(\frac{1}{x}+\frac{1}{2017}-\frac{1}{x}\)
=> \(-\frac{1}{x+1}=\frac{1}{2017}\)
=> \(-1\cdot2017=\left(x+1\right)\cdot1\)
=> \(-2017=x+1\)
=> \(x=-2017-1\)
=> \(x=-2018\)
Vậy \(x=-2018\)
x2 + 2x = 0
=> x(x + 2) = 0
=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
(x - 2) + 3.x2 - 6x = 0
=> (x - 2) + 3x2 - 3x . 2 = 0
=> (x - 2) + 3x.(x - 2) = 0
=> (1 + 3x)(x - 2) = 0
=> \(\orbr{\begin{cases}1+3x=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)
\(\frac{\left(x+1\right)3}{111\cdot3}=\frac{3x+3}{333}\)
\(\frac{\left(y+2\right)2}{222\cdot2}=\frac{2y+4}{444}\)
Ta có: \(\frac{3x+3}{333}=\frac{2y+4}{444}=\frac{z+3}{333}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{3x+3}{333}=\frac{2y+4}{444}=\frac{z+3}{333}=\frac{3x+3+2y+4+z+3}{333+444+333}=\frac{\left(3x+2y+z\right)+\left(3+4+3\right)}{1110}=\frac{989+10}{1110}=\frac{999}{1110}=\frac{9}{10}\)
\(\frac{3x+3}{333}=\frac{9}{10}\Rightarrow3x+3=\frac{2997}{10}\Rightarrow3x=\frac{2967}{10}\Rightarrow x=\frac{989}{10}=98,9\)
Tìm y và z tương tự nhé! Ko hiểu chỗ nào thì nói tớ!
\(\frac{x+5}{100}+\frac{x+5}{99}=\frac{x+5}{98}+\frac{x+5}{97}\)
\(\Leftrightarrow\frac{x+5}{100}+\frac{x+5}{99}-\frac{x+5}{98}-\frac{x+5}{97}=0\)
\(\Leftrightarrow\left(x+5\right)\left(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\right)=0\)
\(\Leftrightarrow x+5=0\) (Vì: \(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\ne0\) )
\(\Leftrightarrow x=-5\)
\(\frac{x+5}{100}+\frac{x+5}{99}=\frac{x+5}{98}+\frac{x+5}{97}\)
\(\Rightarrow\frac{x+5}{100}+\frac{x+5}{99}-\frac{x+5}{98}-\frac{x+5}{97}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\right)=0\)
Mà \(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\ne0\)
\(\Rightarrow x+5=0\)
\(\Rightarrow x=-5\)
Vậy \(x=-5\)
Ta có : \(x^2+1>0\)
Vậy để \(\frac{x^2+1}{x-5}< 0\) thì \(x-5< 0\Rightarrow x< 5\)