Bài `4`

`1, 2y(x+2)-3x-6`

`=2y(x+2) -(3x+6)`

`=2y(x+2) -3(x+2)`

`=(x+2)(2y-3)`

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

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

`=3(x+4) -x(x+4)`

`=(x+3)(3-x)`

`3, 2(x+5) -x^2-5x`

`=2(x+5)-(x^2+5x)`

`=2(x+5)-x(x+5)`

`=(x+5)(2-x)`

`4, x^2 +6x-3(x+6)`

`= (x^2+6x) -3(x+6)`

`=x(x+6)-3(x+6)`

`=(x+6)(x-3)`

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

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

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

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

`6,x(x-y)+2x-2y`

`=x(x-y)+2(x-y)`

`=(x-y)(x+2`

Nửa chu vi hình chữ nhật là :

`96 : 2= 48(m)`

Chiều dài hình chữ nhật là :

`48:(5+3) xx 5=30(m)`

Chiều rộng hình chữ nhật là :

`48-30=18(m)`

Diện tích hình chữ nhật là :

`30 xx 18=540(m^2)`

Diện tích còn lại là :

`540  xx 1/12 = 45(m^2)`

`->B`

dùng máy đt nên khộ hả=)))

\(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\\ =\dfrac{1}{x+2}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}\\ =\dfrac{2}{2x-1}\)

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`x^3+1` chứ cậu nhỉ?

\(\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x-1\right)\left(x^2-x+1\right)}\\ =\dfrac{-3x^2+x+1+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{-2x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{-2\left(x^2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{-2\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{-2\left(x-1\right)}{x^2-x+1}\)

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Nghiệm?

\(\left(x-5\right)^{2016}+\left(x-5\right)^{2018}=0\\ \Rightarrow\left(x-5\right)^{2016}\left[1+\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\1+\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x\in\varnothing\end{matrix}\right.\)