x khac +-3

A=\(\hept{\begin{cases}\\\end{cases}\frac{xI\left(x-3\right)I}{5x^2-45}=\frac{xI\left(x-3\right)I}{5\left(x^2-3^2\right)}}\)

\(\frac{xIx-3I\overline{ }}{5\left(x-3\right)\left(x+3\right)^{ }_{ }}\)

x>3 A=\(\frac{x}{5\left(x+3\right)}\)

x<3 A=-\(\frac{x}{5\left(x+3\right)}\)

Câu 5: B

Câu 6: 

a: ĐKXĐ: \(x-2\ne0\)

=>\(x\ne2\)

b: ĐKXĐ: \(x+1\ne0\)

=>\(x\ne-1\)

8:

\(A=\dfrac{x^2+4}{3x^2-6x}+\dfrac{5x+2}{3x}-\dfrac{4x}{3x^2-6x}\)

\(=\dfrac{x^2+4-4x}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)

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

\(=\dfrac{x-2+5x+2}{3x}=\dfrac{6x}{3x}=2\)

7: 

\(\dfrac{8x^3yz}{24xy^2}\)

\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)

\(=\dfrac{x^2z}{3y}\)

\(\frac{2x^2-3x-20}{x^2-16}\)

\(=\frac{2x^2-8x+5x-20}{\left(x-4\right)\left(x+4\right)}\)

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

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

\(=\frac{2x+5}{x+4}\)

Vậy ...

a kham khảo nha , e nhờ a e lm chứ ko phải e lm nha ! 

\(\left(x-2\right)\left(\frac{3}{x}+2-\frac{5}{2x}-4+\frac{8}{x^2}-4\right)\)

\(\left(x-2\right)\left[\left(\frac{3}{x}-\frac{5}{2x}\right)-6+\frac{8}{x^2}\right]\)

\(\left(x-2\right)\left(\frac{1}{2x}-6+\frac{8}{x^2}\right)\)

\(\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2x-4}+\frac{8}{x^2-4}\right)\)

\(=\left(x-2\right)\left[\frac{3}{x+2}-\frac{5}{2\left(x-2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{3.2\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{8.2}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)-5\left(x+2\right)+16}{2\left(x-2\right)\left(x+2\right)}\right]\)

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

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

\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)

\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)

a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)

\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)

\(=\dfrac{7y}{8x}\)

a, x+45-[90+(-20)+5-(-45)]

=x+45-[90+(-20)+5+45]

=x+45-90+20-5-45

=x+(45-45)-(90+20)-5

=x+0-110-5

=x-115

vay x-115

b, x+(294+13)+(94-13)

=x+307+81

=x+388

vay x+388

đúng rồi

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

\(1,\Leftrightarrow x=10\\ 2,=x^2-4x+4-x^2+4x=4\\ 3,=\left(x+y\right)^2-49=\left(x+y+7\right)\left(x+y-7\right)\)

Câu 1 :

\(2x-20=0\)

\(2x=0+20\)

\(2x=20\)

\(2.x=20\)

\(x=20:2\)

\(x=10\)