Ta có: A = 98 x a + 2 x a + 118 x b - 18 x b

             = (98+2) x a + (118-18) x b

             = 100 x a + 100 x b

             = 100 x (a+b)

             = 100 x 240

             = 24000 

=24000

A=98 x a+118 x b+2 x a-18 x b

=(98+2) x a+(118-18) x b

=100 x a+100 x b

=100 x (a+b)

=100 x 20 x 12

=2000 x 12

=24000

a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)

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

c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)

d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)

`a, (3x^2y)/(2xy^5)`

`= (3x)/(2y^4)`

`b, (3x^2-3x)/(x-1)`

`= (3x(x-1))/(x-1)`

`= 3x`

`c, (ab^2-a^2b)/(2a^2+a)`

`= (b(a-b))/((2a+1))`

`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.

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

\(a,\) Điều kiện xác định: \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)

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

\(=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)

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

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

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

\(=\dfrac{4}{x-3}\)

\(c,x=1\Rightarrow A=\dfrac{4}{1-3}=-2\)

ta có: a) (x+2)(x²-2x+4)-(18+x³)

=(x³+4)-(18+x³)

=x³+8-18-x³

=8-18

=-10

b)(x-3)(x+3)-(x+5)(x-1)=(x²-9)-(x²-x+5x-5)

=(x²-9)-(x²+4x-5)

=x²-9-x²-4x+5

=-9-4x+5

=-4-4x

=4(-1-x)

a) (x+2)(x²-2x+4)-(18+x³)=(x³+4)-(18+x³)

=x³+4-18-x³

=4-18

=-14

b)(x-3)(x+3)-(x+5)(x-1)=(x²-9)-(x²-x+5x-5)

=(x²-9)-(x²+4x-5)

=x²-9-x²-4x+5

=-9-4x+5

=-4-4x

=4(-1-x)

* \(\left(x+2\right)\left(x^2-2x+4\right)-\left(18+x^3\right)\)

\(=x^3+2^3-18-x^3\)

\(=8-18=-10\)

* \(\left(x-3\right)\left(x+3\right)-\left(x+5\right)\left(x-1\right)\)

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

\(=x^2-9-x^2+x-5x+5\)

\(=-4x-4=-4.\left(x+1\right)\)

(Nếu đúng thì k cho mình với nhé nhe!)

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

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

\(A=\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}\)

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

a)

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

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

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

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

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

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

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

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

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

b) Thay \(A=4\) vào phân thức \(A\) , ta có:

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

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

\(x-3=1\)

\(x=1+3\)

\(x=4\)

Vậy \(x=4\) khi \(A=4\)

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

= \(\frac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}=\frac{4}{x-3}\)

b) để A=4 thì \(\frac{4}{x-3}=4\)=> x-3=1=|> x=4

Cây dừa, cây hành, cây ngô