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18 tháng 12 2018

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

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

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

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

18 tháng 12 2018

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 \(A=4\)

16 tháng 6 2021

1) A = \(\dfrac{2x-1}{x+3}\) = \(\dfrac{3}{2}\) (=) (2x-1).2 = 3.(x+3)

                          (=) 4x-2 =3x+9

                          (=) 4x-3x = 9+2

                         (=) x = 11 (tm)

2) Để \(\dfrac{A}{B}\)\(^{x^2}\)+5 (=) \(\dfrac{2x-1}{x+3}\)\(\dfrac{2}{x^2-9}\) <  \(x^2\)+5 

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

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

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

                    (=) \(2x^2\)- 7x + 3 < \(2x^2\)+ 10

                    (=)  (\(2x^2\)-\(2x^2\)) - 7x < -3 +10

                    (=) -7x < 7 

                    (=) x > -1

                   

29 tháng 12 2022

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

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

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

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

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

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

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

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

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

a: \(x^2-9-x^2\left(x^2-9\right)\)

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

\(=\left(x^2-9\right)\left(1-x^2\right)\)

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

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

\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-y^2\right)\)

\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)

c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

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

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

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

d: \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

e: \(3x^2-4x-4\)

\(=3x^2-6x+2x-4\)

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

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

g: \(x^4+64y^4\)

\(=x^4+16x^2y^2+64y^4-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)

 

h: \(a^2+b^2+2a-2b-2ab\)

\(=a^2-2ab+b^2+2a-2b\)

\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)

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

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

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

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

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

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

Câu 1:

b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

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

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

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

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

c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)

Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)

\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)

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

29 tháng 6 2016

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

25 tháng 12 2017

Cây dừa, cây hành, cây ngô

30 tháng 12 2021

1: \(B=\dfrac{6x+x^2-3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x}{x-3}\)

30 tháng 12 2021

Giúp em câu 2 nữa ạ

19 tháng 1 2022

a) -ĐKXĐ của A:

x+3≠0 ⇔x≠-3.

x2-9≠0 ⇔(x-3)(x+3)≠0 ⇔x-3≠0 hay x+3≠0⇔x≠3 hay x≠-3.

x-3≠0 ⇔x≠3.

b) B=x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)

c) A=\(\dfrac{x}{x+3}-\dfrac{6x}{x^2-9}+\dfrac{2}{x-3}\)=\(\dfrac{x\left(x-3\right)+2\left(x+3\right)-6x}{\left(x+3\right)\left(x-3\right)}\)=\(\dfrac{x^2-3x+2x+6-6x}{\left(x+3\right)\left(x-3\right)}\)=\(\dfrac{x^2-7x+6}{x^2-9}\)

d)- Vì x=37 thỏa mãn ĐKXĐ của A và A=\(\dfrac{x^2-7x+6}{x^2-9}\)nên:

A=\(\dfrac{37^2-7.37+6}{37^2-9}=\dfrac{279}{340}\)