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bài 1

 a)   \(7x-12=5x+3\)

    \(\Leftrightarrow7x-5x=3+12\)

     \(\Leftrightarrow2x=15\)

       \(\Leftrightarrow x=7,5\)

vậy phương trình có tập nghiệm là  \(S=\left\{7,5\right\}\)

cây hỏi là gì bn ?

câu hỏi đâu bn

`Answer:`

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1-1-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1=\frac{1-x}{2002}+1-\frac{x}{2003}+1\)

\(\Leftrightarrow\frac{2-x+2001}{2001}=\frac{1-x+2002}{2002}-\frac{x+2003}{2003}\)

\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

\(\Leftrightarrow\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)

\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow2003-x=0\)

\(\Leftrightarrow x=2003\)

`Answer:`

`(1+1/x)^3.(1+x^3)=16`

`<=>(x+1)^3.(1+x^3)=16x^3`

`<=>(x+1)^4.(x^2-x+1)=16x^3`

`<=>x^6+3x^5+3x^4+2x^3+3x^2+3x+1=16x^3`

`<=>x^6+3x^5+3x^4-14x^3+3x^3+3=0`

`<=>(x-1)^2.(x+1)^2.(x^2+x+1)=0`

`<=>x=1`

a; \(\Rightarrow6x-8x+3=8\Leftrightarrow-x=5\Leftrightarrow x=-5\)

b, \(\Rightarrow4x-8-6x+9=12x-12\Leftrightarrow14x=13\Leftrightarrow x=\frac{13}{14}\)

c, TH1 : x = 0 

TH2 : x + 1 = 0 <=> x = -1 

TH3 : x + 3/4 = 0 <=> x = -3/4 

d, \(\Leftrightarrow\left(x^2-1\right)\left(2x-1\right)-\left(x^2-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(2x-1-x-3\right)=0\)

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

TH1 : x = 1 ; TH2 : x = -2 ; TH3 : x = 4 

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