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1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PT <=> \(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\frac{4}{3}x^2\left(\frac{1}{5}x-\frac{2}{3}\right)-\frac{22}{45}x^2-\left(\frac{1}{5}x-\frac{2}{3}\right)=0\)
<=> \(x^2\left(\frac{4x}{15}-\frac{2}{5}-\frac{4x}{15}+\frac{8}{9}-\frac{22}{45}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)=0\)
<=> \(x^2.0-\frac{1}{5}x+\frac{2}{3}=0\)
<=> \(\frac{1}{5}x=\frac{2}{3}\Rightarrow x=\frac{2}{3}:\frac{1}{5}=\frac{10}{3}\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\frac{4x^2}{5}\times\frac{2x-3}{6}-\frac{3x-10}{15}\times\frac{4x^2+3}{3}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{4x^2\left(2x-3\right)}{30}-\frac{\left(3x-10\right)\left(4x^2+3\right)}{45}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{12x^2\left(2x-3\right)}{90}-\frac{2\left(3x-10\right)\left(4x^2+3\right)}{90}=\frac{44x^2}{90}\)
\(\Leftrightarrow12x^2\left(2x-3\right)-2\left(3x-10\right)\left(4x^2+3\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-2\left(12x^3+9x-40x^2-30\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-24x^3-18x+80x^2+60=44x^2\)
\(\Leftrightarrow24x^3-36x^2-24x^3-18x+80x^2-44x^2=-60\)
\(\Leftrightarrow\left(24x^3-24x^3\right)+\left(-36x^2+80x^2-44x^2\right)-18x=-60\)
\(\Leftrightarrow-18x=-60\)
\(\Leftrightarrow x=\frac{-60}{-18}\)
\(\Leftrightarrow x=\frac{10}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ -4x(x - 5) - 2x(8 - 2x) = -3
=> -4x2 + 20x - 16x + 4x2 = -3
=> 4x = -3
=> x = -3/4
b/ \(\frac{x-1}{-15}=-\frac{60}{x-1}\Rightarrow\left(x-1\right)^2=\left(-60\right)\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900\Rightarrow\orbr{\begin{cases}x-1=30\\x-1=-30\end{cases}\Rightarrow\orbr{\begin{cases}x=31\\x=-29\end{cases}}}\)
Vậy x = -29 , x = 31
![](https://rs.olm.vn/images/avt/0.png?1311)
c, Trừ hai vế cho 6
Vế trái thì lấy từng số hạng trừ 1 là được
\(\frac{5}{x+2}-\frac{5}{x+3}=\frac{4x-8}{\left(x+2\right)\left(x+3\right)}\)ĐKXĐ : \(x\ne-2;-3\)
\(\Leftrightarrow\frac{5\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}-\frac{5\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}=\frac{4x-8}{\left(x+2\right)\left(x+3\right)}\)
\(\Rightarrow5x+15-5x-10=4x-8\)
\(\Leftrightarrow4x-8-15+10=0\)
\(\Leftrightarrow4x-13=0\)
\(\Leftrightarrow x=\frac{13}{4}\)( t/m ĐKXĐ )
Vậy....
\(\frac{5}{x+2}-\frac{5}{x+3}=\frac{4x-8}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{5}{x+2}\left(x+2\right)\left(x+3\right)-\frac{5}{x+3}\left(x+2\right)\left(x+3\right)=\frac{4x-8}{\left(x+2\right)\left(x+3\right)}\left(x+2\right)\left(x+3\right)\)
<=> 5(x + 3) - 5(x + 2) = 4x - 8
<=> 5 = 4x - 8
<=> 5 - 5 = 4x - 8 - 5
<=> 0 = 4x - 13
<=> 4x + 0 = 13
<=> 4x = 13
<=> x = 13 : 4
=> x = \(\frac{13}{4}\)