giúp mk vs ạ cảm ơn mn nhiều

Đáp án `+` Giải thích các bước giải `!`

`1.`

`a)`

`85. 12,7+5,3. 12,7`

`= 12,7(85+5,3)`

`= 12,7. 90,3`

`= 1146,81`

`b)`

`52. 143-52. 39-8. 26`

`= 52. 143-52. 39-4. 52`

`= 52(143-39-4)`

`= 52. 100`

`= 5200`

`2.`

`a)`

`5x-20y`

`= 5(x-4y)`

`b)`

`5x(x-1)-3x^2(x-1)`

`= (5x-3x^2)(x-1)`

`= x(5-3x)(x-1)`

`c)`

`x(x+y)-5x-5y`

`= x(x+y)-(5x+5y)`

`= x(x+y)-5(x+y)`

`= (x-5)(x+y)`

`3.`

`a)`

`x+5x^2 = 0`

`<=> x(1+5x) = 0`

`⇔` \(\left[ \begin{array}{l}x=0\\1+5x=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\5x=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x=-\dfrac{1}{5}\end{array} \right.\) 

Vậy `S= {0; -(1)/5}`

`b)`

`x+1 = (x+1)^2`

`<=> x+1 = x^2+2x+1`

`<=> x+1-x^2-2x-1 = 0`

`<=> -x^2-x = 0`

`<=> -x(x+1) = 0`

`<=>` \(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\) 

Vậy `S= {0; -1}`

`c)`

`x^3+x = 0`

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

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

`⇔` \(\left[ \begin{array}{l}x=0\\x^2=-1\end{array} \right.\) 

Vì `x^2 >= 0` `AA x`

Mà `x^2 = -1` (vô lý)

`<=> x = 0`

Vậy `S= {0}`