Đáp án:

5) \(\left[ \begin{array}{l}
x =  - 2\\
x = \dfrac{1}{5}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\left( {4x - 10} \right)\left( {20 + 5x} \right) = 0\\
 \to \left[ \begin{array}{l}
4x - 10 = 0\\
20 + 5x = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - 4
\end{array} \right.\\
2)DK:x \ne  \pm 1\\
\dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2} - 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
 \to {x^2} + 2x + 1 - {x^2} + 2x - 1 - 4 = 0\\
 \to 4x - 4 = 0\\
 \to x = 1\left( l \right)\\
 \to x \in \emptyset \\
3)\dfrac{{4\left( {2x - 8} \right) - 6\left( {3x + 1} \right) - 3\left( {9x - 2} \right) - 2\left( {3x - 1} \right)}}{{24}} = 0\\
 \to 8x - 32 - 18x - 6 - 18x + 6 - 6x + 2 = 0\\
 \to  - 34x - 30 = 0\\
 \to x =  - \dfrac{{15}}{{17}}\\
5)(x + 2)(3 - 4x) = {x^2} + 4x + 4\\
 \to  - 4{x^2} - 5x + 6 = {x^2} + 4x + 4\\
 \to 5{x^2} + 9x - 2 = 0\\
 \to 5{x^2} + 10x - x - 2 = 0\\
 \to 5x\left( {x + 2} \right) - \left( {x + 2} \right) = 0\\
 \to \left( {x + 2} \right)\left( {5x - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x =  - 2\\
x = \dfrac{1}{5}
\end{array} \right.\\
6)(2x + 1){(x + 1)^2}(2x + 3) = 0\\
 \to \left[ \begin{array}{l}
2x + 1 = 0\\
x + 1 = 0\\
2x + 3 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x =  - 1\\
x =  - \dfrac{3}{2}
\end{array} \right.
\end{array}\)

Đáp án:

↓↓↓

Giải thích các bước giải:

`a,(4x-10)(20-5x)=0`

⇒ \(\left[ \begin{array}{l}4x-10=0\\20-5x=0\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}x=5/2\\x=4\end{array} \right.\) 

`b,(x+1)/(x-1)-(x-1)/(x+1)=4/(x^2-1)`

⇒`x`$\neq$ `±1`

⇒`(x+1)/(x-1)-(x-1)/(x+1)-4/(x^2-1)=0`

⇒`((x+1)^2-(x-1)^2-4)/((x-1)(x+1))=0`

⇒`(4(x-1))/((x-1)(x+1))=0`

⇒`4/(x+1)=0`

⇒`4=0`

⇒`x∈∅`

`c,(2x-8)/6-(3x+1)/4=(9x-2)/8+(3x-1)/12`

⇒`4(2x-8)-6(3x+1)=3(9x-2)+2(3x-1)`

⇒`8x-32-18x-6=27x-6+6x-2`

⇒`-43x=30`

⇒`x=-30/43`

`d,(x+2)(3-4x)=x^2+4x+4`

⇒`3x-4x^2+6-8x=x^2+4x+4`

⇒`-5x-4x^2+6=x^2+4x+4`

⇒`-5x-4x^2+6-x^2-4x-4=0`

⇒`(x+2)(5x-1)=0`

⇒\(\left[ \begin{array}{l}x=-2\\x=1/5\end{array} \right.\)

`e,(2x+1)(x+1)^2(2x+3)=0`

⇒\(\left[ \begin{array}{l}2x+1=0\\(x+1)^2=0\\2x+3=0\end{array} \right.\)

⇒\(\left[ \begin{array}{l}x=-1/2\\x=-1\\x=-3/2\end{array} \right.\)