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

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

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: A=x+3+|x-3|

=x+3+3-x(x<=3)

=6

b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)

\(=\left|x+2\right|-\left|x\right|\)

=x+2-x=2

c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)

\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)

Em ơi mình đăng bài sang bên môn toán nha

Bài 1:

a: \(\left(3x-2\right)\left(9x^2+6x+4\right)-3\left(9x^3-2\right)\)

\(=\left(3x\right)^3-2^3-27x^3+6\)

\(=27x^3-8-27x^3+6=-2\)

b: \(\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)

\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)

\(=x^4-16-x^4+9=-16+9=-7\)

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

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

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

Bài 2:

a: \(-4x^2-4x-2\)

\(=-4x^2-4x-1-1\)

\(=-\left(2x+1\right)^2-1\le-1<0\forall x\)

yggucbsgfuyvfbsudy

????????

a: Ta có: \(P=\left(x-1\right)^2-4x\left(x+1\right)\left(x-1\right)+3\)

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

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

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

b: Thay x=-2 vào P, ta được:

\(P=-4\cdot\left(-8\right)+4-4+4=36\)

1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

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

\(=\left(-3\right)^2+5\left(2x-3\right)\)

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

\(A=x^2+4x-21-x^2-4x+5=-16\\ B=-2\left(4x^2+20x+25\right)-\left(1-16x^2\right)\\ B=-8x^2-40x-50-1+16x^2=8x^2-40x-51\\ C=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ D=x^3+1-\left(x^3-1\right)=2\\ E=x^3-3x^2+3x-1-x^3+1-9x^2+1=-12x^2+3x+1\)