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

\(\Leftrightarrow3x^2+3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

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

\(\Rightarrow x^3+3x^2+3x+1-x^3+1-2=0\)

\(\Rightarrow3x^2+3x=0\Rightarrow3x\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(=x^3+3x^2+3x+1-x^3+1-2=0\\ \Leftrightarrow3x^2+3x=0\\ \Leftrightarrow3x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(\frac{3}{4}-\frac{x}{2}-1\frac{1}{2}=\)

\(\frac{3}{4}-\frac{x}{2}-\frac{3}{2}=0\)

\(\frac{3}{4}-\frac{x}{2}=0+\frac{3}{2}\)

\(\frac{3}{4}-\frac{x}{2}=\frac{3}{2}\)

\(\frac{x}{2}=\frac{3}{4}-\frac{3}{2}\)

\(\frac{x}{2}=-\frac{3}{4}\)

\(x:2=-\frac{3}{4}\)

\(x=-\frac{3}{4}.2\)

\(x=-\frac{3}{2}\)

=> x = -3

TÌM X À

Đunga rồi >< mình nhầm. Đây toán lớp 7 ><

BEAR CHỊU

= -1

giải thì tự xử lí viết ra dài  nản

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

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

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`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

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

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

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`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

1,5 - 9/4 x+ 1/2x=0

9/4x+1/2x= -1/5

(9/4+1/2)x= -1/5

11/4x=-1/5

x= -4/55

 Bây giờ tớ sẽ sửa:

1,5 - 9/4x+1/2x=0

-9/4x+1/2x= -1,5

(-9/4 + 1/2) x= -1,5

-7/4x = -1,5

x= 6/7

1.

\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\\Delta=\left(m+1\right)^2-4m\left(m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\-3m^2+7m+1< 0\end{matrix}\right.\)

\(\Leftrightarrow m< \dfrac{7-\sqrt{61}}{6}\)

2.

\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\\Delta'=4\left(m+1\right)^2-m\left(m-5\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\3m^2+13m+4\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\-4\le m\le-\dfrac{1}{3}\end{matrix}\right.\)

Không tồn tại m thỏa mãn