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Bài 1:
a) \(\frac{2x+1}{3}-\frac{x}{4}=2\)
\(\Leftrightarrow\frac{4\left(2x+1\right)}{12}-\frac{3x}{12}-\frac{24}{12}=0\)
\(\Leftrightarrow8x+4-3x-24=0\)
\(\Leftrightarrow5x-20=0\)
\(\Leftrightarrow5x=20\)
\(\Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
b) \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\)
ĐKXĐ: \(x\ne0;x\ne-5\)
\(\Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x\left(2x+5\right)}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2-5x=0\)
\(\Leftrightarrow10x+25=0\)
\(\Leftrightarrow10x=-25\)
\(\Leftrightarrow x=-\frac{5}{2}\left(TM\right)\)
Vậy \(S=\left\{-\frac{5}{2}\right\}\)
#Học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
@Nguyễn Lê Phước Thịnh bạn có thể chỉ chỗ mình sai sót được không ạ? Mình mò không ra ._.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)
\(Đkxđ:\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
\(Pt\Leftrightarrow x\left(x+2\right)-2=x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tmđk\right)\end{matrix}\right.\)
Vậy .........
\(b,Đkxđ:x\ne-5\)
Ta có: \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow2x-5=3\left(x+5\right)\)
\(\Leftrightarrow x=20\left(tmđk\right)\)
Vậy .........
c, \(Đkxđ:x\ne3\)
Ta có: \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktmđk\right)\end{matrix}\right.\)
Vậy ............
![](https://rs.olm.vn/images/avt/0.png?1311)
mk chỉ giải đc có bài 1 thui nha bn
\(\frac{4}{x-2}+\frac{1}{x+3}=0\)
ĐKXĐ: x ≠ 2 và x ≠ -3
QĐKM:
⇔(x+3)4 + (x-2)1 = 0
⇔4x + 12 + x - 2 = 0
⇔4x + x = -12 + 2
⇔5x = -10
⇔x= -2
S={-2}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5 - (x - 6) = 4(3 - 2x)
<=> 5 - x + 6 = 12 - 8x
<=> -x + 8x = 12 - 11
<=> 7x = 1
<=> x = 1/7
Vậy S = {1/7}
b) 2x(x - 3) + 5(x - 3) = 0
<=> (2x + 5)(x - 3) = 0
<=> \(\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)
Vậy S = {-5/2; 3}
c)ĐK: x \(\ne\)1; x \(\ne\)2
\(\frac{3x-5}{x-2}-\frac{2x-5}{x-1}=1\)
<=> \(\frac{\left(3x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(2x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\)
<=> 3x2 - 8x + 5 - 2x2 + 9x - 10 = x2 - 3x + 2
<=> x2 + x - 5 = x2 - 3x + 2
<=> x2 + x - x2 + 3x = 2 + 5
<=> 4x = 7
<=> x = 7/4
Vậy S = {7/4}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2x-5}{x-1}=\frac{1-3x}{x+1}\)(1)
\(DKXD:\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)Ta có:
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x+1\right)=\left(x-1\right)\left(1-3x\right)\)
\(\Leftrightarrow2x^2+2x-5x-5=x-3x^2-1+3x\)
\(\Leftrightarrow2x^2+2x-5x-5-x+3x^2+1-3x=0\)
\(\Leftrightarrow5x^2-7x-4=0\)
\(\Leftrightarrow5\left(x^2-\frac{7}{5}x\right)-4=0\)
\(\Leftrightarrow5\left(x^2-2.x.\frac{7}{10}+\frac{49}{100}\right)-5.\frac{49}{100}-4=0\)
\(\Leftrightarrow5\left(x^2-\frac{7}{10}\right)^2-\frac{129}{20}=0\)
\(\Leftrightarrow5\left(x^2-\frac{7}{10}\right)^2=\frac{129}{20}\)
\(\Leftrightarrow\left(x-\frac{7}{10}\right)^2=\frac{129}{100}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{10}=\sqrt{\frac{129}{100}}=\frac{\sqrt{129}}{10}\\x-\frac{7}{10}=-\sqrt{\frac{129}{100}}=-\frac{\sqrt{129}}{10}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{129}}{10}+\frac{7}{10}\\x=-\frac{\sqrt{129}}{10}+\frac{7}{10}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{129}+7}{10}\\x=\frac{7-\sqrt{129}}{10}\end{cases}}}\)
\(\frac{2x-5}{x-1}=\frac{1-3x}{x+1}\)ĐKXĐ: \(x\ne+-1\)
\(\Rightarrow\left(2x-5\right)\left(x+1\right)=\left(x-1\right)\left(1-3x\right)\)
\(\Leftrightarrow2x^2-5x+2x-5=x-3x^2+3x-1\)
\(\Leftrightarrow2x^2+3x^2-5x+2x+x-3x-5+1=0\)
\(\Leftrightarrow5x^2-5x-4=0\)
......
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\) \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)
\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
\(ĐKXĐ:x\ne2\)
\(pt\Leftrightarrow2x-5>x-2\)
\(\Leftrightarrow2x-x>5-2\)
\(\Leftrightarrow x>3\)
Vậy x > 3
\(\frac{2x-5}{2-x}>-1\)
=> \(2x-5>-\left(2-x\right)\)
=> \(2x-5>x-2\)
=> \(2x-x>5-2\)
=> \(x>3\)