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\(\Leftrightarrow\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{x\left(x+1\right)}=-1\left(đkxđ:x\ne\pm1;0;2;3\right)\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{x+1}=-1\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x+1}=-1\)
\(\Leftrightarrow\dfrac{4}{x^2-2x-3}=-1\)
\(\Leftrightarrow x^2-2x-3=-4\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\left(loai\right)\)
Vậy không có giá trị x thỏa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(^{\dfrac{-x^2-xy}{5\left(x^2-y^2\right)}}\).\(\dfrac{3\left(x^3-y^3\right)}{x^2-xy}\)
=\(\dfrac{-3\left(x-y\right)}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính M = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)
\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)
\(\Leftrightarrow x^2-x-72=0\)
=>(x-9)(x+8)=0
=>x=9 hoặc x=-8
6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0 hoặc x=5
5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x
<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8
6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9
<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5
7, <=> (x-1)^2 = (3x+3)^2
<=> (x-1-3x-3)(x-1+3x+3) = 0
<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2
8, = (x^2-10x-15)(x^2-10x+25)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{4x}{x^2-4x+7}-1\right)+\left(\dfrac{3x}{x^2-5x+7}-1\right)=2\)
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ĐKXĐ: \(x\ne\pm3,x\ne\dfrac{9}{2}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{2x-9}.\dfrac{3\left(x-3\right)-x}{x\left(x-3\right)}\right]\) : \(\dfrac{x^2-5x-6}{-2\left(x-3\right)\left(x+3\right)}\)
= \(\left[\dfrac{x}{2\left(x-3\right)}-\dfrac{x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x-3}\right]:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x\left(x+3\right)-2x^2+2\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}:\dfrac{-\left(x^2-5x-6\right)}{2\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}{-2\left(x^2-5x-6\right)\left(x-3\right)\left(x+3\right)}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+144+5x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow-2x=51\)
hay \(x=-\dfrac{51}{2}\)
Vậy: \(x=-\dfrac{51}{2}\)
4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)
\(\Leftrightarrow2x+2-x+2=6x-3\)
\(\Leftrightarrow x+4-6x+3=0\)
\(\Leftrightarrow-5x+7=0\)
\(\Leftrightarrow-5x=-7\)
hay \(x=\dfrac{7}{5}\)
Vậy: \(x=\dfrac{7}{5}\)
1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(2\left(5x-2\right)=3\left(5-3x\right)\)
\(10x-4=15-9x\)
\(10x+9x=15+4\)
\(19x=19\)
\(x=1\)
Vậy \(x=1\)
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1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{2x-9-\left(x+3\right)\left(x-3\right)+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
\(P=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}\left(x\ne2;x\ne3\right)\)
\(=\dfrac{2x-9}{x^2-2x-3x+6}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\dfrac{2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-x^2+9+2x^2-3x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-2x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x\left(x-2\right)+\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+1}{x-3}\)
#Urushi
\(=\dfrac{2}{\left(x-2\right)\left(x-3\right)}-\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6-2x+4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\dfrac{10}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\)