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\(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne4\end{cases}}\)
\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)+\left(x-1\right)\left(x-4\right)+2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
\(a,\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\left(x\ne1;x\ne-2\right)\)
\(\Leftrightarrow\frac{3}{x^2+x-2}-\frac{1}{x-1}+\frac{7}{x+2}=0\)
\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{1\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{7x-7}{\left(x-1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3-x-2+7x-7}{\left(x-1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{6x-8}{\left(x-1\right)\left(x+2\right)}=0\)
=> 6x-8=0
<=> x=\(\frac{8}{6}=\frac{4}{3}\left(tmđk\right)\)
b) ĐKXĐ: x khác 2; x khác 4
\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\)
<=> \(\frac{2}{\left(x-2\right)\left(x-4\right)}+\frac{x-1}{x-2}=\frac{x+3}{x-4}\)
<=> 2(x - 2) + (x - 1)(x - 4)(x - 2) = (x + 3)(x - 2)(x - 2)
<=> x^3 - 7x^2 + 16x - 12 = -x^3 + x^2 + 8x - 12
<=> x^2 - 7x^2 + 16x - 12 + x^3 - x^2 + 8x - 12 = 0
<=> 2x^3 - 8x^2 + 8x = 0
<=> 2x(x - 2)(x - 2) = 0
<=> 2x = 0 hoặc x - 2 = 0
<=> x = 0 (tmđk) hoặc x = 2 (ktmđk)
=> x = 2
ĐKXĐ: \(x\ne-2;-3;-4\)
Ta có: \(x+\frac{x}{x+2}+\frac{x+3}{x^2+5x+6}+\frac{x+4}{x^2+6x+8}=1\)
<=> \(\frac{x\left(x+2\right)}{x+2}+\frac{x}{x+2}+\frac{x+3}{\left(x+2\right)\left(x+3\right)}+\frac{x+4}{\left(x+2\right)\left(x+4\right)}\)=1
<=> \(\frac{x^2+2x}{x+2}+\frac{x}{x+2}+\frac{1}{x+2}+\frac{1}{x+2}=1\)
<=> \(\frac{x^2+3x+2}{x+2}=1\)<=>\(\frac{\left(x+1\right)\left(x+2\right)}{x+2}=1\)<=>x+1=1
<=>x=0
Vậy x=0
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
\(-537x^2+5054x=-541x^2+5092x\)
\(-537x^2+5054x+541x^2-5092x=0\)
\(4x^2-38x=0\)
\(x\left(2x-19\right)=0\)
\(\orbr{\begin{cases}x=0\\2x=19\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{19}{2}\end{cases}}\)
Ta có
\(-x^2+6x-8=-\left(x^2-6x+8\right)=-\left(x^2-2x-4x+8\right)=-\left[x\left(x-2\right)-4\left(x-2\right)\right]\)
\(=-\left(x-2\right)\left(x-4\right)\)
MTC:\(\left(x-2\right)\left(x-4\right)\)
\(\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{\left(x-2\right)\left(x-4\right)}\)
\(\frac{x^2-x-4x+4+x^2+3x-2x-6}{\left(x-2\right)\left(x-4\right)}=-\frac{2}{\left(x-2\right)\left(x-4\right)}\)
\(\frac{2x^2-4x-2}{\left(x-2\right)\left(x-4\right)}=-\frac{2}{\left(x-2\right)\left(x-4\right)}\)
\(2x^2-4x-2+2=0\Rightarrow2x^2-4x=0\Rightarrow2x\left(x-4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\Rightarrow x=0\\x-4=0\Rightarrow x=4\end{cases}}\)
a) trước nha
T i c k cho mình nha bạn cảm ơn sẽ làm típ câu b)
b) x3 - 8 - (x - 2) (x + 2) = 0
=> x3 - (2)3 - x2 - 22 = 0
mk làm bậy thui!!! mới lên lớp 8 mà!! 6756886787696969768658585685685685858978467
1) (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 = 4x2 - 8
<=> 12x + 9 = -8
<=> 12x = -17
<=> x = 17/12
1) (2x - 3)^2 = 4x^2 - 8
<=> 4x^2 - 12x + 9 = 4x^2 - 8
<=> 4x^2 - 12x + 9 - 4x^2 = -8
<=> -12x + 9 = -8
<=> -12x = -8 - 9
<=> -12x = -17
<=> x = 17/12
2) x - (x + 2)(x - 3) = 4 - x^2
<=> x - x^2 + 3x - 2x + 6 = 4 - x^2
<=> 2x - x^2 + 6 = 4 - x^2
<=> 2x - x^2 + 6 + x^2 = 4
<=> 2x + 6 = 4
<=> 2x = 4 + 6
<=> 2x = 10
<=> x = 5
3) 3x - (x - 3)(x + 1) = 6x - x^2
<=> 3x - x^2 - x + 3x + 3 = 6x - x^2
<=> 5x - x^2 + 3 = 6x - x^2
<=> 5x - x^2 + 3 + x^2 = 6x
<=> 5x + 3 = 6x
<=> 3 = 6x - 5x
<=> 3 = x
4) 3x/4 = 6
<=> 3x = 6.4
<=> 3x = 24
<=> x = 8
5) 7 + 5x/3 = x - 2
<=> 21 + 5x = 3x - 6
<=> 5x = 3x - 6 - 21
<=> 5x = 3x - 27
<=> 5x - 3x = -27
<=> 2x = -27
<=> x = -27/2
6) x + 4 = 2/5x - 3
<=> 5x + 20 = 2x - 15
<=> 5x + 20 - 2x = -15
<=> 3x + 20 = -15
<=> 3x = -15 - 20
<=> 3x = -35
<=> x = -35/3
7) 1 + x/9 = 4/3
<=> x/9 = 4/3 - 1
<=> x/9 = 1/3
<=> x = 3
A . 3x + 2(x + 1) = 6x - 7
<=> 3x + 2x + 2 = 6x -7
<=> 5x - 6x = -7 - 2
<=> -x = -9
<=> x =9
B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)
=> 3(x +3) < 5(5 -x)
<=> 3x+9 < 25 - 5x
<=> 3x + 5x < 25 - 9
<=> 8x < 16
<=> x < 2
C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)
<=> 5(x - 4) + 2x = 2(x +1)
<=> 5x - 20 + 2x = 2x + 2
<=>7x - 2x = 2 + 20
<=> 5x = 22
<=> x =\(\frac{22}{5}\)
ĐKXĐ: x≠4; x≠2
Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{x^2-6x+8}=0\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)+\left(x-1\right)\left(x-4\right)+2=0\)
\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
Vì 2≠0 nên
\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\)
Vậy: x=0