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a, Ta có: \(a\left(ax-1\right)=x-1\)
\(\Leftrightarrow a^2x-a=x-1\)
\(\Leftrightarrow a^2x-x=a-1\)
\(\Leftrightarrow x\left(a-1\right)\left(a+1\right)=a-1\)
Với \(a\ne\pm1\)=> Pt có nghiệm duy nhất \(x=\frac{a-1}{a+1}\)
Với \(a=1\)=> Pt có nghiệm đúng với mọi x
Với \(a=-1\)=> Pt vô nghiệm
a) ĐKXĐ : \(x\ne\pm a\).
Với \(a=-3\) khi đó ta có pt :
\(A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left(-9+1\right)}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(-3-x\right)}{\left(-3-x\right)\left(-3+x\right)}+\frac{24}{\left(-3-x\right)\left(-3+x\right)}=0\)
\(\Rightarrow x^2-9-\left(-3x-x^2-9-3x\right)+24=0\)
\(\Leftrightarrow2x^2+6x+24=0\)
\(\Leftrightarrow x^2+3x+12=0\) ( vô nghiệm )
Phần b) tương tự.
\(A=\frac{x+a}{a-x}-\frac{x-a}{a+x}=\frac{a\left(3x+1\right)}{a^2-x^2}\)
\(=\frac{x+a}{a-x}+\frac{x-a}{a+x}=\frac{a\left(3+1\right)}{\left(a-x\right)\left(a+x\right)}\)
\(=\frac{\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+1\right)}=\frac{a\left(3a+1\right)}{\left(a+x\right)\left(a-x\right)}\)
\(\Leftrightarrow\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)=a\left(3a+1\right)\)
\(\Leftrightarrow x^2+2ax+a^2-ax-x^2-a^2+ax=3a^2+a\)
\(\Leftrightarrow2ax=3a^2+a\)
\(\Leftrightarrow x=\frac{3a^2+a}{2a}\left(a\ne0\right)\)
a) Khi x=-3 => \(x=\frac{3\cdot\left(-3\right)^2-3}{2\left(-3\right)}=-13\)
b) a=1
\(\Leftrightarrow x=\frac{3\cdot1^2+1}{2\cdot1}=2\)
a) \(ĐKXĐ:x\ne\pm3\)
Với a = -3
\(\Leftrightarrow A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}=\frac{24}{9-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}+\frac{24}{x^2-9}=0\)
\(\Leftrightarrow\frac{-\left(x-3\right)^2-\left(x+3\right)^2+24}{x^2-9}=0\)
\(\Leftrightarrow-x^2+6x-9-x^2-6x-9+24=0\)
\(\Leftrightarrow-2x^2+6=0\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\pm\sqrt{3}\)(tm)
Vậy với \(a=-3\Leftrightarrow x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
b) \(ĐKXĐ:x\ne\pm1\)
Với a = 1
\(\Leftrightarrow A=\frac{x+1}{1-x}-\frac{x-1}{1+x}=\frac{3+1}{1-x^2}\)
\(\Leftrightarrow\frac{x+1}{1-x}-\frac{x-1}{1+x}+\frac{4}{x^2-1}=0\)
\(\Leftrightarrow\frac{-\left(x+1\right)^2-\left(x-1\right)^2+4}{x^2-1}=0\)
\(\Leftrightarrow-x^2-2x-1-x^2+2x-1+4=0\)
\(\Leftrightarrow-2x^2+2=0\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)(ktm)
Vậy với \(a=1\Leftrightarrow x\in\varnothing\)
c) \(ĐKXĐ:a\ne\pm\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào phương trình, ta đươc :
\(A=\frac{\frac{1}{2}+a}{a-\frac{1}{2}}-\frac{\frac{1}{2}-a}{a+\frac{1}{2}}=\frac{a\left(3a+1\right)}{a^2-\frac{1}{4}}\)
\(\Leftrightarrow\frac{a+\frac{1}{2}}{a-\frac{1}{2}}+\frac{a-\frac{1}{2}}{a+\frac{1}{2}}-\frac{3a^2+a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow\frac{\left(a+\frac{1}{2}\right)^2+\left(a-\frac{1}{2}\right)^2-3a^2-a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow a^2+a+\frac{1}{4}+a^2-a+\frac{1}{4}-3a^2-a=0\)
\(\Leftrightarrow-a^2-a+\frac{1}{2}=0\)
\(\Leftrightarrow a^2+a-\frac{1}{2}=0\)
\(\Leftrightarrow\left(a+\frac{1}{2}\right)^2-\frac{3}{4}=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}\\a=-\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{-\sqrt{3}-1}{2}\end{cases}}\)(TM)
Vậy với \(x=\frac{1}{2}\Leftrightarrow a\in\left\{\frac{\sqrt{3}-1}{2};\frac{-\sqrt{3}-1}{2}\right\}\)
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