a) Tam thức \(2{x^2} + 3x + m + 1\) có \(\Delta  = {3^2} - 4.2.\left( {m + 1} \right) = 1 - 8m\)

Vì \(a = 2 > 0\) nên để \(2{x^2} + 3x + m + 1 > 0\) với mọi \(x \in \mathbb{R}\) khi và chỉ khi \(\Delta  < 0 \Leftrightarrow 1 - 8m < 0 \Leftrightarrow m > \frac{1}{8}\)

Vậy khi \(m > \frac{1}{8}\) thì \(2{x^2} + 3x + m + 1 > 0\) với mọi \(x \in \mathbb{R}\)

b) Tam thức \(m{x^2} + 5x - 3\) có \(\Delta  = {5^2} - 4.m.\left( { - 3} \right) = 25 + 12m\)

Đề \(m{x^2} + 5x - 3 \le 0\) với mọi \(x \in \mathbb{R}\) khi và chỉ khi \(m < 0\) và \(\Delta  = 25 + 12m \le 0 \Leftrightarrow m \le  - \frac{{25}}{{12}}\)

Vậy \(m{x^2} + 5x - 3 \le 0\) với mọi \(x \in \mathbb{R}\) khi \(m \le  - \frac{{25}}{{12}}\)

a)
A=0
\(x\left(x-\dfrac{4}{9}\right)=0\)
x=0 hoặc x-4/9=0
x=0 hoặc x=4/9
b)
A>0
\(x\left(x-\dfrac{4}{9}\right)>0\)
TH1
x>0  và x-4/9 >0
x>0  và x>4/9
TH2
x<0 và x-4/9<0
x<0 và x<4/9
c)
A<0
\(x\left(x-\dfrac{4}{9}\right)< 0\)
TH1
x<0 và x-4/9>0
x<0 và x>4/9
TH2
x>0 và x-4/9 <0
x>0 và x<4/9

a) A = \(\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\left(\frac{1}{1-x}-1\right)\)

A = \(\frac{3x^2+3x-3}{x^2+2x-x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\left(\frac{1-1+x}{1-x}\right)\)

A = \(\frac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\frac{x}{1-x}\)

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

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

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

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

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

A = \(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

A = \(\frac{x+1}{x-1}\) (Đk: \(x-1\ge0\) => x \(\ge\)1)

b) Ta có: A = \(\frac{x+1}{x-1}=\frac{\left(x-1\right)+2}{x-1}=1+\frac{2}{x-1}\)

Để A \(\in\)Z <=> 2 \(⋮\)x - 1

<=> x - 1 \(\in\)Ư(2) = {1; -1; 2; -2}

<=> x \(\in\){2; 0; 3; -1}

c) Ta có: A < 0

=> \(\frac{x+1}{x-1}< 0\)

=> \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)

=> \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(loại) hoặc \(\hept{\begin{cases}x>-1\\x< 1\end{cases}}\) 

=> -1 < x < 1

Edogawa Conan

Thiếu dòng đầu  \(ĐKXĐ:\hept{\begin{cases}x\ne1\\x\ne-2\\x\ne0\end{cases}}\)

ĐKXĐ : \(\) x # +1 ; x # - 1 ; x # -2 ; x # 0 ; x # 2

 Ta có: \(A=\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}.\left(\frac{1}{1-x}-1\right)\)

  \(=\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}.\frac{x}{1-x}\)

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

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

  \(=\frac{3x^2+3x-3}{x^2+x-2}-\frac{2x^2-5}{x^2+x-2}\)

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

   \(\frac{x+1}{x-1}\)

b. Ta có:  \(A=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)

Để A nhận giá trị nguyên thì: \(2⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(2\right)\)

  +) x - 1 = 1 => x = 2   (loại)

  +) x - 1 = 2 => x = 3  

  +) x - 1 = -1 => x = 0  (loại)

  +) x - 1 = -2 => x = -1    (loại)

Vậy x = 3 là giá trị cần tìm.

c.  \(A< 0\Leftrightarrow\frac{x+1}{x-1}< 0\)

\(\Leftrightarrow\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)   hoặc \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>-1\\x< 1\end{cases}}\)    hoặc   \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(vô lý)

Vậy \(-1< x< 1\) và x # 0 là giá trị cần tìm

\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)

\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)

\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)

\(P>0\Leftrightarrow x>-4\)

sai lớp :>>>

a) \(ĐKXĐ:\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}\)

     \(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)

          \(=\left[\frac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right]:\frac{2\left(1-2x\right)}{x+1}-\frac{3x+1-x^2}{3x}\)

       \(=\frac{\left(x+1\right)\left(x+2\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

       \(=\frac{2-8x^2}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

      \(=\frac{1+2x-3x-1+x^2}{3x}\)

      \(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)

b)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)

  \(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)

Dựa theo kết quả câu  a)   mk lm tiếp câu  b)   nhé:

b)  ĐKXĐ:  \(x\ne0;\)\(x\ne-1;\)\(x\ne0,5\)

 \(A< 0\) thì   \(\frac{x-1}{3}< 0\)

  \(\Leftrightarrow\)\(x-1< 0\)   (do  \(3>0\))

\(\Leftrightarrow\)\(x< 1\)

Vậy  với    \(x< 1\)thỏa mãn  ĐKXĐ   thì   \(A< 0\)

a) \(-ĐKXĐ:x\ne\pm2;1\)

Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)

\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)

b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)

\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)

Vậy với mọi x thỏa mãn x>1 thì A > 0

c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)

\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy x = -1;-2

\(A=\dfrac{-\left(x^2+2xy+y^2\right)+4x^2+4xy+y^2}{x^2+2xy+y^2}=-1+\left(\dfrac{2x+y}{x+y}\right)^2\ge-1\)

\(A_{min}=-1\) khi \(2x+y=0\)

\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)

\(A=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-3x}\right)\)

\(=\frac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x^2-3x}{x^2-3x}\)

\(=\frac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2x-3\right)}{x\left(x-3\right)}\)

\(=\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x-3}{x-3}\)

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

b

Xét hơi bị nhiều TH nhá:(

Để \(A>0\) thì \(\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}>0\)

TH1:\(4x\left(2x-3\right)>0;\left(2+x\right)\left(x-3\right)>0\)

\(TH2:4x\left(2x-3\right)< 0;\left(2+x\right)\left(x-3\right)< 0\)

Bạn tự xét nốt nhá!

c

\(\left|x-7\right|=4\Rightarrow x-7=4;x-7=-4\)

\(\Rightarrow x=11;x=3\)

Thay vào .....