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\(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)\(ĐKXĐ:x\ne1;2;3;4\)
\(\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)
\(\Leftrightarrow\left(\frac{\left(x-1\right)^2}{x-1}+\frac{1}{x-1}\right)+\left(\frac{\left(x-4\right)^2}{x-4}+\frac{4}{x-4}\right)=\left(\frac{\left(x-2\right)^2}{x-2}+\frac{2}{x-2}\right)+\left(\frac{\left(x-3\right)^2}{x-3}+\frac{3}{x-3}\right)\)
\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{1}{x-4}=x-2+\frac{1}{x-2}+x-3+\frac{1}{x-3}\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{x-4+4x-4}{\left(x-1\right)\left(x-4\right)}=\frac{2x-6+3x-6}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)
\(\Leftrightarrow\left(5x-8\right)\left(x^2-5x+6\right)=\left(5x-12\right)\left(x^2-5x+4\right)\)
Tự giải ra rồi tìm x nhé
dấu suy ra số 4 là 1/(x+1) + 1/(x+4) mà.
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
ĐK \(x\ne\left\{1;2;3;4\right\}\)
Ta có \(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)
\(\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)
\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{4}{x-4}=x-2+\frac{2}{x-2}+x-3+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)\(\Leftrightarrow\left(5x-8\right)\left(x^2-5x+6\right)=\left(5x-12\right)\left(x^2-5x+4\right)\)
\(\Leftrightarrow5x^3-25x^2+30x-8x^2+40x-48=5x^3-25x^2+20x-12x^2+60x-48\)
\(\Leftrightarrow4x^2-10x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}\left(tm\right)}\)
Vậy x=0 hoặc x=5/2
ĐKXĐ: \(x\ne-1,-2,-3,-4\)
\(\Leftrightarrow\frac{\left(x+1\right)^2+1}{x+1}+\frac{\left(x+4\right)^2+4}{x+4}=\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\frac{1}{x+1}+x+4+\frac{4}{x+4}=x+2+\frac{2}{x+2}+x+3+\frac{3}{x+3}\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{x+4}=\frac{1}{x+2}+\frac{1}{x+3}\)
\(\Leftrightarrow\frac{x}{x+1}+\frac{x}{x+4}=\frac{x}{x+2}+\frac{x}{x+3}\)
\(\Leftrightarrow x\left(\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}\right)=0\)
\(\Leftrightarrow x\left(\frac{1}{x^2+3x+2}-\frac{1}{x^2+7x+12}\right)=0\)
\(\Leftrightarrow-x\left(\frac{4x+10}{\left(x^2+3x+2\right)\left(x^2+7x+12\right)}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{5}{2}\end{cases}}\)Thỏa mãn ĐKXĐ
Ta có Pt
<=>\(\frac{\left(x+1\right)^2+1}{x+1}+\frac{\left(x+4\right)^2+4}{x+4}=\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+3\right)^2+3}{x+3}\)
<=>\(x+1+\frac{1}{x+1}+x+4+\frac{4}{x+4}=x+2+\frac{2}{x+2}+x+3+\frac{3}{x+3}\)
<=>\(\frac{1}{x+1}+\frac{4}{x+4}=\frac{2}{x+2}+\frac{3}{x+3}\)
<=>\(1-\frac{1}{x+1}+1-\frac{4}{x+4}=1-\frac{2}{x+2}+1-\frac{3}{x+3}\)
<=>\(\frac{x}{x+1}+\frac{x}{x+4}=\frac{x}{x+2}+\frac{x}{x+3}\Leftrightarrow x\left(\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}\right)=0\)
<=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}=0\left(1\right)\end{cases}}\)
Giải pt (1) , ta có
\(\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}-\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}=0\)
<=>\(\frac{1}{x^2+3x+2}-\frac{1}{x^2+7x+12}=0\Leftrightarrow x^2+3x+2=x^2+7x+12\)
<=>\(4x+10=0\Leftrightarrow x=-\frac{5}{2}\)
nhớ đối chiếu đk nhé !
^_^