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![](https://rs.olm.vn/images/avt/0.png?1311)
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
![](https://rs.olm.vn/images/avt/0.png?1311)
c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)
=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)
=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12
=>-16x-1=-15x+12
=>-x=13
=>x=-13
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân tích : x2-3x +2=(x-1)(x-2) , x2-4x +3 = (x-1 )(x-3) , điều kiện : x # 1, x # 2 ,x # 3
pt tương đương với : \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}=\frac{2x+5+x+1}{\left(x-1\right)\left(x-3\right)}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}=\frac{3\left(x+2\right)}{\left(x-1\right)\left(x-3\right)}\)
<=> \(\frac{\left(x+4\right)\left(x-3\right)-3\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
<=> \(\frac{x\left(1-2x\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
<=> x=0 hoặc x=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
\(ĐKXĐ:x\ne1;5;9\)
\(pt\Leftrightarrow\frac{2x-1}{\left(x-1\right)\left(x-5\right)}+\frac{\left(x-2\right)}{\left(x-1\right)\left(x-9\right)}=\frac{3x-12}{\left(x-9\right)\left(x+5\right)}\)
\(\Rightarrow\left(2x-1\right)\left(x-9\right)+\left(x-2\right)\left(x-9\right)=\left(3x-12\right)\left(x-1\right)\)
\(=>2x^2-x-18x+9+x^2-2x+5x-10=3x^2-12-3x+12\)
\(=>3x^2-16x-1=3x^2-15x+12\)
=>x=-13