a, ĐKXĐ : \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)hoặc \(2x^2-x^3\ne0\)hay \(x\ne\pm2;0\)

\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

\(=\left(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)

\(=\frac{-x^2-2x-1-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}:\frac{x-3}{x\left(2-x\right)}\)

\(=\frac{-4x^2-6x+3}{\left(x-2\right)\left(x+2\right)}.\frac{-x\left(x-2\right)}{x-3}=\frac{\left(-4x^2-6x+3\right)\left(-x\right)}{\left(x+2\right)\left(x-3\right)}=\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}\)

b, Ta có : A > 0 hay \(\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}>0\)

\(\Leftrightarrow x\left(4x^2+6x-3\right)>0\)

\(\Leftrightarrow4x^2+6x-3>0\) bạn xem lại bài mình có chỗ nào sai ko nhé !!! 

c, Ta có : \(\left|x-7\right|=4\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)

TH1 : Thay x = 11 vào phân thức trên : ... 

TH2 : Thay x = 3 vào phân thức trên : .... tự làm 

\(\left(3x+5\right)\left(x+2\right)=x^2-5x\)

\(\Leftrightarrow3x^2+6x+5x+10=x^2-5x\)

\(\Leftrightarrow3x^2+11x+10-x^2+5x=0\)

\(\Leftrightarrow2x^2+16x+10=0\)

\(\Leftrightarrow2\left(x^2+8x+5\ne0\right)=0\)

Vậy phương trình vô nghiệm 

\(2x\left(3x-1\right)=3x-1\)

\(\Leftrightarrow2x=\frac{3x-1}{3x-1}\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

\(2x\left(3x-1\right)=3x-1\)

\(\Leftrightarrow2x\left(3x-1\right)-\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x-1\right)=0\Leftrightarrow x=\frac{1}{3};\frac{1}{2}\)

x⁶ - 2x⁴ + x³ + x² - x 

= x⁶ - x⁴ - x⁴ + x³ + x² - x

= ( x⁶ - x⁴ ) - ( x⁴ - x² ) + ( x³ - x )

= x³( x³ - x ) - x( x³ - x ) + ( x³ - x )

= ( x³ - x )( x³ - x + 1 )

= 6.( 6 + 1 ) = 42

ta có x=2 vì 2³-2=6

=>2⁶-2.2⁴+2³+2²-2

=64-2.16+8+4-2

=64-32+8+4-2

=32+8+4-2

=40+4-2

=44-2

=42

a³ + b³ + c³ = 3abc

⇒ a³ + b³ + c³ - 3abc = 0

⇒ ( a³ + b³ ) + c³ - 3abc = 0

⇒ ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

⇒ [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

⇒ ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

⇒ ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

⇒ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Vì a,b,c > 0 ⇒ a + b + c > 0 ⇒ a + b + c = 0 không xảy ra

Xét a² + b² + c² - ab - bc - ac = 0

⇒ 2( a² + b² + c² - ab - bc - ac ) = 2.0

⇒ 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

⇒ ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - ac + c² ) = 0

⇒ ( a - b )² + ( b - c )² + ( a - c )² = 0

Vì \(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(a-c\right)^2\end{cases}}\ge0\forall a,b,c\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

Dấu "=" xảy ra khi a = b = c

Khi đó : a - b + 1/29 = a - a + 1/29 = 1/29

Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

Vì \(a,b,c>0\)nên \(a+b+c>0\)suy ra \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\)

Ta có: \(6x^2-x-2=0\)

   \(\Leftrightarrow6x^2-4x+3x-2=0\)

   \(\Leftrightarrow2x.\left(3x-2\right)+\left(3x-2\right)=0\)

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

   \(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\3x-2=0\end{cases}}\)

    \(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)

6x² - 2 - x = 0

=> 6x² + 3x - 4x - 2 = 0

=> 3x(2x + 1) - 2(2x + 1) = 0

=> (3x - 2)(2x + 1) = 0

=> \(\orbr{\begin{cases}3x-2=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\2x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{1}{2}\end{cases}}\)

( x - 1 )( x² + 5x - 2 ) - ( x³ - 1 ) = 0

<=> ( x - 1 )( x² + 5x - 2 ) - ( x - 1 )( x² + x + 1 ) = 0

<=> ( x - 1 )( x² + 5x - 2 - x² -  x - 1 ) = 0

<=> ( x - 1 )( 4x - 3 ) = 0

<=> x - 1 = 0 hoặc 4x - 3 = 0

<=> x = 1 hoặc x = 3/4

Vậy pt có tập nghiệm S = { 1 ; 3/4 }

\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\Leftrightarrow x=1;\frac{3}{4}\)