a) Đề phải là: \(A=\left(x-2\right)\left(x+2\right)-\left(x-1\right)\left(x^2+2x+1\right)-x^2\left(4-x\right)\) chứ bạn

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

           \(=x^2-4-x^3+1-4x^2+x^3\) 

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

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

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

              \(\Leftrightarrow x^2=-1\)

Vì \(x^2\ge0\left(\forall x\right)\) \(\Rightarrow x\in\varnothing\)

Vậy x vô nghiệm nếu A có giá trị bằng 0

P/s: không chắc lắm

đề sao cũng đúng mà

a)  \(A=\left(x-2\right)\left(x+2\right)-\left(x-1\right)\left(x^2-2x+1\right)-x^2\left(4-x\right)\)

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

=> \(A=x^2-4-x^3+3x^2-3x+1-4x^2+x^3\)

=> \(A=-3x-3\)

b)  Cho A=0

=> \(A=-3x-3=0\)

=> \(-3x=3\)

=> \(x=-1\)

ĐKXĐ: x\(\ne\)1, x\(\ne\)-1

MTC (x-1)(x+1)

\(\Leftrightarrow\)(\(\frac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)+ \(\frac{2\left(x-1\right)}{MTC}\)-\(\frac{-\left(5-x\right)}{MTC}\)) : \(\frac{1-2x}{MTC}\)

\(\Rightarrow\)\(\left[-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)\right]:\left(1-2x\right)\)

\(\Leftrightarrow\frac{-x-1+2x-2+5-x}{1-2x}\)

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

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

b. mình chỉ biết  \(x< \frac{1}{2}\) thôi chứ ko biết làm sao

hình như là giải Bất phương trình \(\frac{2}{1-2x}>0\)

a, Rút gọn :

\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x-10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\frac{1\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}-\frac{2x-10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\frac{x-5+2x+10-2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\frac{x+15}{\left(x+5\right)\left(x-5\right)}\)

3 phút trước (13:18)

Kb đi buồn quá

ê pn sao hay thế

Cho biểu thức : A= ( 3/2x+4 + x/2-x + 2x^2+3/x^2-4 ) : (2x-1/4x-8)

a.Rút gọn A

b.Tìm giá trị của A biết |x - 1| = 3

c.Tìm x để A < 2

d.Tìm x để A = |1|

\(a,P=\dfrac{2x^2-1}{x^2+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\left(x\ne0;x\ne-1\right)\\ P=\dfrac{2x^2-1-x^2+1+3x}{x\left(x+1\right)}=\dfrac{x\left(x+3\right)}{x\left(x+1\right)}=\dfrac{x+3}{x+1}\\ b,P=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\left(tm\right)\\ c,x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=1\left(x\ne0\right)\\ \Leftrightarrow P=\dfrac{1+3}{1+1}=\dfrac{4}{2}=2\)

1,

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

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

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

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

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

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

\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)

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

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

a,\(ĐKXĐ:\hept{\begin{cases}x\ne\mp2\\x\ne3\\x\ne0\end{cases}}\)

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

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

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

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

A= \(\frac{3\left(x-3\right)+\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{18}{\left(9-x^2\right)}\)

A= \(\frac{3x-9+x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{x^2-9}\)

A=\(\frac{3x+x-9+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)

A=\(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)

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

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

để A=4

=>   \(\frac{4}{x-3}=4\)

<=>  x-3=1

<=> x=4

a, Rút gọn : 

\(A=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

\(A=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{1\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}\)

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

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

b, Để A = 4 

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

\(\Leftrightarrow4\left(x-3\right)=4\)

\(\Leftrightarrow4x-12=4\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\)

Vậy để a = 4 thì x = 4