Bài 4: Bất phương trình bậc nhất một ẩn.

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

Suy ra: \(\left(x+1\right)^2\cdot\left(x^2-x+1\right)-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=2\left(x+2\right)^2\)

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

\(\Leftrightarrow x^4+x^3+x+1-x^4+x^3+x-1=2\left(x+2\right)^2\)

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

\(\Leftrightarrow2x^2\left(x+1\right)-2\left(x+2\right)^2=0\)

\(1,ĐK:x\ne\pm2;x\ne5\\ 2,P=\dfrac{2}{x-2}-\dfrac{2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\\ P=\dfrac{1}{x-2}-\dfrac{2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2-2x}{\left(x-2\right)\left(x+2\right)}\\ P=\dfrac{2-x}{\left(x-2\right)\left(x+2\right)}=\dfrac{-1}{x+2}\\ 3,P\in Z\Leftrightarrow x+2\inƯ\left(-1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-3;-1\right\}\left(tm\right)\)

1) \(ĐKXĐ:x\ne2;x\ne-2;x\ne5.\)

2) \(P=\dfrac{2x-10}{x^2-7x+10}-\dfrac{2x}{x^2-4}+\dfrac{1}{2-x}\left(ĐKXĐ:x\ne2;x\ne-2;x\ne5\right).\)

\(P=\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x-2\right)}-\dfrac{2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}.\)

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

\(P=\dfrac{2\left(x+2\right)-2x-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}.\)

\(P=\dfrac{2x+4-2x-x-2}{\left(x-2\right)\left(x+2\right)}.\)

\(P=\dfrac{-x+2}{\left(x-2\right)\left(x+2\right)}.\)

\(P=\dfrac{-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}.\)

\(P=\dfrac{-1}{x+2}.\)

3) Để \(P\in Z\) <=> \(\dfrac{-1}{x+2}\in Z\) => \(x+2\inƯ\left(-1\right)\) <=>\(x+2\in\left\{-1;1\right\}\).

TH1: x + 2 = -1 <=> x = -3 (TM).

Th2: x + 2 = 1 <=> x = -1 (TM).

Vậy \(x\in\left\{-3;-1\right\}\).

\(\dfrac{2x-3}{x-1}< \dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow6x-9< x-1\Leftrightarrow5x< 8\Leftrightarrow x< \dfrac{8}{5}\) và ĐK \(x\ne1\)

\(\dfrac{2x-3}{x-1}>\dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow x-1< 6x-9\Leftrightarrow5x>8\Leftrightarrow x>\dfrac{8}{5}\) và ĐK \(x\ne1\)

<=> 3x>9

<=> x>3

Ta có: \(7x-3>4x+6\)

\(\Leftrightarrow3x>9\)

hay x>3

- Đặt \(f\left(x\right)=\dfrac{2x-3}{19+8x}\)

- Lập bảng xét dấu :

- Từ bảng xét dấu : - Để : \(f\left(x\right)< 0\)

\(\Leftrightarrow-\dfrac{19}{8}< x< \dfrac{3}{2}\)

Vậy ...

Ta có: \(\dfrac{2x-3}{8x+19}< 0\)

Trường hợp 1: \(\left\{{}\begin{matrix}2x-3>0\\8x+19< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{19}{8}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Trường hợp 2: \(\left\{{}\begin{matrix}2x-3< 0\\8x+19>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x>-\dfrac{19}{8}\end{matrix}\right.\Leftrightarrow-\dfrac{19}{8}< x< \dfrac{3}{2}\)

Vậy: S={x|\(-\dfrac{19}{8}< x< \dfrac{3}{2}\)}

\(x^2-2x+8< 0\)

\(\Leftrightarrow x^2-2x+1+7< 0\)

\(\Leftrightarrow\left(x-1\right)^2+7< 0\)

PTVN.

`x^2 - 2x + 8 < 0`

`<=> (x-1)^2 + 7 < 0`

`<=> (x-1)^2 < -7`

Vì `(x-1)^2 > -7` với mọi `x`

`=>` vô nghiệm 

Vậy `x \in RR`

Vì `x^2 >=0`

`=> x^2 + 20 >= 20 > 0 forall x`

`=> x^2+20>0 forall x`.

Phân tích cái gì vậy bạn:v?

\(x^2+20>0\)

ta có \(x^2\)≥0=>\(x^2+20>20\) (20>0)

=>x∈R pt vô số nghiệm

bạn chụp lại giúp mik nhé

Chụp hơi mờ nha bạn