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1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
a: \(\dfrac{3x-7}{2}+\dfrac{x-1}{3}=-16\)
\(\Leftrightarrow3\left(3x-7\right)+2\left(x-1\right)=-96\)
\(\Leftrightarrow9x-21+2x-2=-96\)
=>11x=-73
hay x=-73/11
b: \(x-\dfrac{x-1}{3}=\dfrac{2x+1}{5}\)
=>15x-5(x-1)=3(2x+1)
=>15x-5x+5=6x+3
=>10x+5=6x+3
=>4x=-2
hay x=-1/2
c: \(\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)
=>14x-7-15x-6=21(x+13)
=>21x+273=-x-13
=>22x=-286
hay x=13
\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
a: \(\Leftrightarrow6x^2-2x=6x^2-13\)
=>-2x=-13
hay x=13/2
b: \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)
=>2x-2x-1=x-6x
=>-5x=-1
hay x=1/5
c: \(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=x^2+3\)
\(\Leftrightarrow x^2+3-x^2-2x-1+x^2-2x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=3