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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)
\(\frac{2x}{5}+\frac{3-2x}{3}\ge\frac{3x+2}{2}\)
\(\Leftrightarrow\)\(\frac{12x}{30}+\frac{10\left(3-2x\right)}{30}\ge\frac{15\left(3x+2\right)}{30}\)
\(\Leftrightarrow\)12x + 30 - 20x \(\ge\) 45x + 30
\(\Leftrightarrow\) 12x - 20x - 45x \(\ge\) -30 + 30
\(\Leftrightarrow\)- 53x \(\ge\)0
\(\Leftrightarrow\)x \(\le\)0
Vậy bất phương trình có nghiệm là : x \(\le0\)
b) \(1-\frac{2x-5}{6}>\frac{3-x}{4}\)
\(\Leftrightarrow\)\(\frac{12}{12}-\frac{2\left(2x-5\right)}{12}>\frac{3\left(3-x\right)}{12}\)
\(\Leftrightarrow\) 12 - 4x + 10 > 9 - 3x
\(\Leftrightarrow\)-4x + 3x > -12 - 10 + 9
\(\Leftrightarrow\)-x > -13
\(\Leftrightarrow\)x < 13
Vậy bất phương trình có nghiệm là : x < 13
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)
\(\frac{12x}{12}-\frac{2\left(5x+2\right)}{12}=\frac{3\left(7-3x\right)}{12}\)
\(12x-10x-4=21-9x\)
\(11x=25\)
\(x=\frac{24}{11}\)
\(b,\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\frac{10x+3}{12}=\frac{15+8x}{9}\)
\(9\left(10x+3\right)=12\left(15+8x\right)\)
\(3\left(10x+3\right)=4\left(8x+15\right)\)
\(30x+9=32x+60\)
\(-2x=51\)
\(x=-\frac{51}{2}\)
\(c,\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\frac{2x}{6}-\frac{3\left(2x+1\right)}{6}=\frac{x-6x}{6}\)
\(2x-6x-3=x-6x\)
\(x=3\)
P/s: Bn xem lại đề bài phần d nha!
=.= hk tốt!!
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)
(đk:x khác \(\frac{1}{2}\))
\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)
Vậy x=\(\frac{25}{7}\)
b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)
(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))
\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)
Vậy x=4
2)
\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)
\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)
\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
Đặt \(x^2-2x+3=t\left(t\ge2\right)\), khi đó phương trình trở thành:
\(\frac{1}{t-1}+\frac{2}{t}=\frac{6}{t+1}\)
\(\Leftrightarrow\frac{t\left(t+1\right)+t^2-1}{\left(t-1\right)t\left(t+1\right)}=\frac{6t\left(t-1\right)}{\left(t-1\right)t\left(t+1\right)}\)
\(\Leftrightarrow t\left(t+1\right)+t^2-1=6t\left(t-1\right)\)
\(\Leftrightarrow2t^2+t-1=6t^2-6t\)
\(\Leftrightarrow-4t^2+7t-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{7+\sqrt{33}}{8}\\t=\frac{7-\sqrt{33}}{8}\end{cases}}\left(ktmđk\right)\)
Vậy phương trình vô nghiệm.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3