1+2+3+......+x= 55
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\(C=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x}{x-1}-\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)\)
\(=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-1\right)\left(x+3\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{3x^2-x+2-x^2-3x-x^2+2x-1}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}=\frac{x-1}{x+3}\)
Vậy C=\(\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}\left(-1\right)=0\)
\(\Rightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{-2}{5}=0\)
\(\Leftrightarrow x\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(\Rightarrow x\times\frac{11}{15}=\frac{2}{5}\)
\(\Rightarrow x=\frac{6}{11}\)
|3x+5|-2=0`
`|3x+5|=2`
TH1: `3x+5>=0 <=>x>=-5/3`
`3x+5=2`
`3x=-3`
`x=-1` (TM)
TH2: `x<-5/3`
`-3x-5=2`
`-3x=7`
`x=-7/3` (L)
Vậy `x=-1`.
\(\left|3x+5\right|-2=0\)
\(\left|3x+5\right|=0+2\)
\(\left|3x+5\right|=2\)
=>\(3x+5=2\) hoặc \(3x+5=-2\)
\(3x=-3\) \(3x=-7\)
\(x=-1\) \(x=\dfrac{-7}{3}\)
X-x/3=5+2/4
3x/3-x/3=20/4+2/4
3x-x/3=22/4
2x/3=11/2
4x/6=33/6
4x=33
x=33/4
Vay...
\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=7\\x-3=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-4\end{cases}}\)
Vậy \(x\in\left\{-4;10\right\}\)
|x - 3| = 7
Xét 2 trường hợp:
TH1: x - 3 = 7
x = 7 + 3
x = 10
TH2: x - 3 = -7
x = -7 + 3
x = -4
Vậy: ...
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{7}=\frac{1}{6}\\x+\frac{2}{7}=\frac{-1}{6}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{42}\\x=\frac{-19}{42}\end{cases}}\)
Vậy ...
Ta có: \(\left|x+\frac{2}{7}\right|=\frac{1}{6}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{2}{7}=\frac{1}{6}\\x+\frac{2}{7}=-\frac{1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}-\frac{2}{7}=-\frac{5}{42}\\x=-\frac{1}{6}-\frac{2}{7}=-\frac{19}{42}\end{cases}}\)
Vậy \(x\in\left\{-\frac{5}{42};-\frac{19}{42}\right\}\)
~Study well~
Áp dụng công thức tính dãy số ta có
x . ( x + 1) : 2 = 55
x . ( x + 1 ) = 110
=> x = 10
từ 1 đến 9 có (9-1)/1+1=9 số
tổng từ 1 đến 9 (9+1)*9/2=45
55-45=10, vậy x = 10