Giải phương trình:
90/x - 36/x-6 =2
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\(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)
\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\) (do \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\))
\(\Leftrightarrow x=100\)
\(\Leftrightarrow36\left(x+6\right)+36\left(x-6\right)=\dfrac{9}{2}\left(x^2-36\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{9}{2}-162-72x=0\)
\(\Leftrightarrow9x^2-144x-324=0\)
\(\Leftrightarrow x^2-16x-36=0\)
=>(x-18)(x+2)=0
=>x=18 hoặc x=-2
ĐKXĐ:\(x\ne\pm6\)
\(\dfrac{36}{x-6}+\dfrac{36}{x+6}=4,5\\ \Leftrightarrow36\left(\dfrac{1}{x-6}+\dfrac{1}{x+6}\right)=4,5\\ \Leftrightarrow\dfrac{x+6}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{\left(x-6\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{x+6+x-6}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{2x}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow x^2-36=16x\\ \Leftrightarrow x^2-16x-36=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(18x+36\right)=0\\ \Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-18\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=18\left(tm\right)\end{matrix}\right.\)
\(\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5\)
\(\Leftrightarrow36\left(x-6\right)+36\left(x+6\right)=4,5\left(x^2-36\right)\)
\(\Leftrightarrow36x-216+36x+216=4,5x^2-162\)
\(\Leftrightarrow-4,5x^2+72x+162=0\)
\(\Leftrightarrow\left(x-18\right)\left(-4,5x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)
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
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) ( 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........
\(\frac{36}{x+6}+\frac{36}{x-6}=\) \(4,5\)\(\left(ĐKCĐ:x\ne\pm6\right)\)
\(\Leftrightarrow\frac{36\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}+\frac{36\left(x+6\right)}{\left(x+6\right)\left(x-6\right)}\)\(=\frac{4,5\left(x-6\right)\left(x+6\right)}{\left(x-6\right)\left(x+6\right)}\)
\(\Leftrightarrow\frac{36x-216}{\left(x-6\right)\left(x+6\right)}+\frac{36x+216}{\left(x-6\right)\left(x+6\right)}\)\(=\frac{4,5x^2-162}{\left(x-6\right)\left(x+6\right)}\)
\(\Rightarrow36x-216+36x+216=4,5x^2-162\)
( đến đây giải phương trình ra rồi đối chiếu đkxđ là xong )
\(\frac{36}{x+6}+\frac{36}{x-6}=4,5\)
\(\frac{36}{x+6}+\frac{36}{x-6}=\frac{4,5\left(x+6\right)\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)
\(DKXD:\hept{\begin{cases}x+6\ne0\\x-6\ne0\\\left(x+6\right)\left(x-6\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-6\\x\ne6\end{cases}}\)
\(\frac{72x}{\left(x+6\right)\left(x-6\right)}=\frac{4,5\left(x+6\right)\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)
\(4,5x^2+72x-162=0\)
\(4,5x^2-9x+81x-162=0\)
\(4,5\left(x-2\right)+81\left(x-2\right)=0\)
\(\left(x-2\right)\left(4,5x-81\right)=0\)
\(\left(x-2\right)4,5\left(x-18\right)=0\)
\(\hept{\begin{cases}x-2=0\\x-18=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\x=18\end{cases}}\)
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 \(x=3\left(nhận\right)\)
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)
90/x - 36/(x - 6) = 2
=>90/x - 36/(x - 6) - 2=0
=>90(x-6)/x(x-6)-36x/x(x-6)-2x(x-6)/x(x-6)=0
=>[90(x-6)-36x-2x(x-6)] / [x(x-6)]=0
=>(90x-540-36x-2x^2+12x)/ [x(x-6)]=0
=>(-2x^2+66x-540)/ [x(x-6)]=0 (*) xác định
Điều kiện để (*) xác định : x(x-6)≠0=>x≠0 hoặc x-6≠0=>x≠0 hoặc x≠6
(*)=>-2x^2+66x-540=0=>x^2-33x+270=0
=>(x^2-15x)-(18x-270)=0
=>x(x-15)-18(x-15)=0
=>(x-18)(x-15)=0
=>(x-18)=0 hoặc (x-15)=0
=>x=18 hoặc x=15 (thỏa mãn điều kiện)
Vậy pt có nghiệm x=18 hoặc x=15
\(\frac{90}{x}-\frac{36}{x-6}=2\) ĐKXĐ : \(x\ne0;x\ne6\)
\(\Leftrightarrow\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=\frac{2x\left(x-6\right)}{x\left(x-6\right)}\)
\(\Leftrightarrow90x-540-36x=2x^2-12x\)
\(\Leftrightarrow66x-2x^2-540=0\)
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