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\(\Leftrightarrow36\left(x+6\right)+36\left(x-6\right)=\dfrac{9}{2}\left(x^2-36\right)\)

\(\Leftrightarrow x^2\cdot\dfrac{9}{2}-162-72x=0\)

\(\Leftrightarrow9x^2-144x-324=0\)

\(\Leftrightarrow x^2-16x-36=0\)

=>(x-18)(x+2)=0

=>x=18 hoặc x=-2

23 tháng 2 2022

ĐKXĐ:\(x\ne\pm6\)

\(\dfrac{36}{x-6}+\dfrac{36}{x+6}=4,5\\ \Leftrightarrow36\left(\dfrac{1}{x-6}+\dfrac{1}{x+6}\right)=4,5\\ \Leftrightarrow\dfrac{x+6}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{\left(x-6\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{x+6+x-6}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{2x}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow x^2-36=16x\\ \Leftrightarrow x^2-16x-36=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(18x+36\right)=0\\ \Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-18\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=18\left(tm\right)\end{matrix}\right.\)

30 tháng 4 2021

\(\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5\)

\(\Leftrightarrow36\left(x-6\right)+36\left(x+6\right)=4,5\left(x^2-36\right)\)

\(\Leftrightarrow36x-216+36x+216=4,5x^2-162\)

\(\Leftrightarrow-4,5x^2+72x+162=0\)

\(\Leftrightarrow\left(x-18\right)\left(-4,5x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)

1 tháng 5 2021

bạn làm rõ hơn ở chỗ này đc ko, mk ko hiểu

⇔−4,5x2+72x+162=0⇔−4,5x2+72x+162=0

⇔(x−18)(−4,5x−9)=0

15 tháng 3 2018

\(\frac{36}{x+6}+\frac{36}{x-6}=\) \(4,5\)\(\left(ĐKCĐ:x\ne\pm6\right)\)

\(\Leftrightarrow\frac{36\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}+\frac{36\left(x+6\right)}{\left(x+6\right)\left(x-6\right)}\)\(=\frac{4,5\left(x-6\right)\left(x+6\right)}{\left(x-6\right)\left(x+6\right)}\)

\(\Leftrightarrow\frac{36x-216}{\left(x-6\right)\left(x+6\right)}+\frac{36x+216}{\left(x-6\right)\left(x+6\right)}\)\(=\frac{4,5x^2-162}{\left(x-6\right)\left(x+6\right)}\)

\(\Rightarrow36x-216+36x+216=4,5x^2-162\)

( đến đây giải phương trình ra rồi đối chiếu đkxđ là xong )

15 tháng 3 2018

\(\frac{36}{x+6}+\frac{36}{x-6}=4,5\)

\(\frac{36}{x+6}+\frac{36}{x-6}=\frac{4,5\left(x+6\right)\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)

\(DKXD:\hept{\begin{cases}x+6\ne0\\x-6\ne0\\\left(x+6\right)\left(x-6\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-6\\x\ne6\end{cases}}\)

\(\frac{72x}{\left(x+6\right)\left(x-6\right)}=\frac{4,5\left(x+6\right)\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)

\(4,5x^2+72x-162=0\)

\(4,5x^2-9x+81x-162=0\)

\(4,5\left(x-2\right)+81\left(x-2\right)=0\)

\(\left(x-2\right)\left(4,5x-81\right)=0\)

\(\left(x-2\right)4,5\left(x-18\right)=0\)

\(\hept{\begin{cases}x-2=0\\x-18=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\x=18\end{cases}}\)

16 tháng 6 2018

\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) ( x # 0 ; x # 6)

\(\dfrac{90\left(x-6\right)-36x}{x\left(x-6\right)}=\dfrac{2x\left(x-6\right)}{x\left(x-6\right)}\)

⇔ 90x - 540 - 36x = 2x2 - 12x

⇔-2x2 + 66x - 540 = 0

⇔ -2( x2 - 33x +270 ) = 0

⇔ x2 - 18x - 15x + 270 = 0

⇔ x( x - 18) - 15( x - 18) = 0

⇔ ( x - 18)( x - 15) = 0

⇔ x = 18 ( TM) hoac x = 15 ( TM)

KL........

1: Ta có: \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)

Suy ra: \(x^2+4x+4+2x-4=x^2\)

\(\Leftrightarrow6x=0\)

hay \(x=0\left(nhận\right)\)

2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)

Suy ra: \(x+6-2x+12=3x+6\)

\(\Leftrightarrow-x-3x=6-18=-12\)

hay \(x=3\left(nhận\right)\)

AH
Akai Haruma
Giáo viên
19 tháng 8 2021

Lời giải:
1. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)

\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)

\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)

2. ĐKXĐ: $x\neq \pm 6$

PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)

\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)

\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)

 

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

2 tháng 5 2017

\(\dfrac{36}{x}+\dfrac{36}{x-12}=\dfrac{9}{2}\)

\(\Rightarrow72\left(x-12\right)+72x\left(x-12\right)-x\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-12\right)\left(72+72x-x\right)=0\)

\(\Leftrightarrow\left(x-12\right)\left(72+71x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\72+71x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=\dfrac{-72}{71}\end{matrix}\right.\)

Vậy phương trình có nghiệm x = 12 hoặc x = \(\dfrac{-72}{71}\) .

30 tháng 3 2018

sai rồi bạn

2 tháng 11 2017

ta có:

(\(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)

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

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

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

= \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\). \(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}\)- \(\dfrac{x}{x-6}\)

= \(\dfrac{6-x}{x-6}\)

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

= -1

19 tháng 12 2018

a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)

\(\Rightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\left(x+3\right)^2-\left(x-3\right)^2=36\)

\(\Rightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=36\)

\(\Rightarrow x^2+6x+9-x^2+6x-9=36\)

\(\Rightarrow12x=36\)

\(\Rightarrow x=\dfrac{36}{12}\)

Vậy x = 3

b) \(x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\)

\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

c) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+17}{15}\)

\(\Rightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)

\(\Rightarrow\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)

\(\Rightarrow\dfrac{6x-3-5x+10}{15}=\dfrac{x+17}{15}\)

... Phần còn lại cũng tương tự như vậy thôi

22 tháng 2 2019

\(\dfrac{x-90}{10}+\dfrac{x-76}{12}=\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)

\(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)=\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)\)\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}=\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-\dfrac{1}{17}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(S=\left\{100\right\}\)