a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a) \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)

ĐK: x≠1

<=>\(\frac{5x-2}{2\left(1-x\right)}+\frac{2x-1}{2}\frac{x^2+x-3}{1-x}=1\)

<=>\(\frac{5x-2+\left(1-x\right).\left(2x-1\right)+2\left(x^2+x-3\right)}{2\left(1-x\right)}=1\)

<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)

<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)

<=> 10x-9=2(1-x)

<=>10x-9=2-2x

<=> 10x+2x= 2+9

<=> 12x=11

<=> x= \(\frac{11}{12}\left(tm\right)\)

b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)

ĐK: x≠2, x≠-2

<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)

<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x²-2x+1)=0

<=> -(6x²-x+12x-2)+9x²+4x-18x-8-3x²+2x-1 = 0

<=> -6x²-11x+2+9x²+4x-18x-8-3x²+2x-1=0

<=> -23x-7=0

<=> -23x=7

<=> x= \(\frac{-7}{23}\left(tm\right)\)

tham khảo câu d trong

https://hoc24.vn/hoi-dap/question/919967.html

c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)

⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

⇒x²+x+1+2x²-5=4x-4

⇔3x²-3x=0

⇔3x(x-1)=0

⇔x=0 (TMĐK) hoặc x=1 (loại)

Vậy tập nghiệm của phương trình đã cho là:S={0}

a.ĐK: 2x²+1\(\ne0\) \(\forall x\)

Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x²+1 >0 với mọi x)

\(\Leftrightarrow x=2\) (TM)

Vậy ...

b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)

Để phương trình bằng 0 thì x²-x-6=0 (Vì x-3\(\ne0\))

\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)

Vậy ...

c. ĐK: x\(\ne\)2

\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)

\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\)

\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)

\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)

Vậy ...

d. ĐK: \(x\ne\pm\frac{1}{3}\)

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)

\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))

\(\Leftrightarrow x=-2\:\left(TM\right)\)

Vậy...

       \(2x-2=8-3x\)

\(\Leftrightarrow\)\(2x+3x=8+2\)

\(\Leftrightarrow\)\(5x=10\)

\(\Leftrightarrow\)\(x=2\)

Vậy...

         \(x^2-3x+1=x+x^2\)

\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)

\(\Leftrightarrow\)\(-4x=-1\)

\(\Leftrightarrow\)\(x=\frac{1}{4}\)

Vậy...

mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))

\(2x-2=8-3x\)

\(\Leftrightarrow\)\(2x+3x=8+2\)

\(\Leftrightarrow\)\(5x=10\)

\(\Leftrightarrow\)\(x=2\)

Vậy \(x=2\)

\(x^2-3x+1=x+x^2\)

\(\Leftrightarrow\)\(4x-1=\left(x^2+x\right)-\left(x^2+x\right)\)

\(\Leftrightarrow\)\(4x-1=0\)

\(\Leftrightarrow\)\(4x=1\)

\(\Leftrightarrow\)\(x=\frac{1}{4}\)

Vậy \(x=\frac{1}{4}\)

\(\left(x^2+1\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2+1=0\\2x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\2x=-4\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{-1}\\x=\frac{-4}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\left\{\varnothing\right\}\\x=-2\end{cases}}}\)

Vậy \(x=-2\)

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