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\(a,\frac{3x+2}{5x+7}=\frac{3x-1}{5x-1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x-1\right)}=\frac{3}{8};\frac{3x+2}{5x+7}=\frac{3}{8}\Leftrightarrow24x+16=15x+21\Leftrightarrow9x=5\Leftrightarrow x=\frac{5}{9}\) \(b,\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow37.7-7x=3x+39\Leftrightarrow259-7x=3x+39\Leftrightarrow220-7x=3x\Leftrightarrow10x=220\Leftrightarrow x=22\) \(c,\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{x+4}{2x+6}=\frac{\left(x+4\right)-\left(x+1\right)}{2x+6-\left(2x+1\right)}=\frac{3}{5};\frac{x+1}{2x+1}=\frac{3}{5}\Leftrightarrow5x+5=6x+3\Leftrightarrow x=2\) \(d,\frac{x-2}{x+2}=\frac{x+3}{x-4}=\frac{\left(x+3\right)-\left(x-2\right)}{\left(x-4\right)-\left(x+2\right)}=\frac{5}{-6};\frac{x-2}{x+2}=\frac{5}{-6}\Leftrightarrow6\left(2-x\right)=5x+10\Leftrightarrow2-6x=5x\Leftrightarrow x=\frac{2}{11}\) \(f,\frac{3x-5}{x}=\frac{9x}{3x+2}=\frac{9x-15}{3x}=\frac{9x-\left(9x-15\right)}{\left(3x+2\right)-3x}=\frac{15}{2};\frac{9x}{3x+2}=\frac{15}{2}\Leftrightarrow18x=45x+30\Leftrightarrow27x+30=0\Leftrightarrow x=\frac{-10}{9}\) \(e,\frac{x+2}{6}=\frac{5x-1}{5}\Leftrightarrow5\left(x+2\right)=6\left(5x-1\right)\Leftrightarrow5x+10=30x-6\Leftrightarrow10=25x-6\Leftrightarrow25x=16\Leftrightarrow x=\frac{16}{25}\)
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\(\frac{5x-1}{3}=\frac{7y-6}{5}=\frac{5x-1-7y+6}{3-5}=\frac{5x-7y+5}{-2}=\frac{5x-7y-7}{4x}\)
\(\frac{5x-7y+5}{-2}=\frac{5x-7y-7}{4x}=\frac{5x-7y+5-5x+7y+7}{-2-4x}=\frac{12}{-2-4x}\)
\(\Rightarrow\frac{5x-1}{3}=\frac{6}{-1-2x}\)
Giải ra tìm x thế vào PT đầu tiên để tìm y
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a) \(5x+\frac{1}{2}-x=\frac{2}{3}\)
\(\left(5x-x\right)+\frac{1}{2}=\frac{2}{3}\)
\(4x+\frac{1}{2}=\frac{2}{3}\)
\(4x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
\(x=\frac{1}{6}\div4\)
\(x=\frac{1}{24}\)
b) \(\frac{-2}{3}x+\frac{3}{7}+\frac{1}{2}x=\frac{-5}{6}\)
\(\left(\frac{-2}{3}x+\frac{1}{2}x\right)+\frac{3}{7}=\frac{-5}{6}\)
\(\frac{-1}{6}x+\frac{3}{7}=\frac{-5}{6}\)
\(\frac{-1}{6}x=\frac{-5}{6}-\frac{3}{7}\)
\(\frac{-1}{6}x=\frac{-53}{42}\)
\(x=\frac{-53}{42}\div\frac{-1}{6}=\frac{53}{7}\)
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\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=3\left(x+6\right)\)
\(\Rightarrow x^2+x+2x+2=3x+18\)
\(\Rightarrow x^2+x+2x-3x=18-2\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
các phần còn lại tương tự :)
a)\(\frac{x+2}{x+6}\) =\(\frac{3}{x+1}\)
<=>\(\frac{\left(x+2\right)\left(x+1\right)}{\left(x+6\right)\left(x+1\right)}\) =\(\frac{3\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}\)
=> ( x+2) ( x+1) = 3(x+6)
<=> x2 +3x +3 = 3x +18
<=> x2 +3x -3x = 18 -3
<=> x2 = 15
=> x = \(\sqrt{15}\)
Vậy x=\(\sqrt{15}\)
b)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x - 7)x+1 - (x - 7)x+1 = 0
<=> 0 = 0
Vậy phương trình có nghiệm với mọi x thuộc R
b/ Chi cần áp dụng tính chất dãy tỷ số bằng nhau thì ra thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\frac{x-1}{x-5}=\frac{6}{7};\left(x-1\right).7=\left(x-5\right).6\)
7x - 7 = 6x - 30
=> 7x - 6x = -30 - (-7)
x = -23
2) \(\frac{x-1}{3}=\frac{x+3}{5};\left(x-1\right).5=\left(x+3\right).3\)
5x - 5 = 3x + 9
=> 5x - 3x = 9 - (-5)
2x = 14
x = 7
3) \(\frac{3}{7}=\frac{2x+1}{3x+5};\left(3x+5\right).3=\left(2x+1\right).7\)
9x + 15 = 14x + 7
9x - 14x = 7-15
5x = -8
x = -8/5
1) =>\(\hept{\begin{cases}x-1=6\\x-5=7\end{cases}=>\hept{\begin{cases}x=6+1=7\\x=7+5=13\end{cases}}}\)
Vậy x\(\varepsilon\){7;13}
2)
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\(\frac{4}{7}=\frac{7}{x^2}\)
\(\Leftrightarrow4x^2=7.7\)
\(\Leftrightarrow\left(2x\right)^2=49\)
\(\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}\)
\(\frac{x-1}{6}=\frac{-7}{5x+3}\)
\(\Rightarrow\left(x-1\right)\left(5x+3\right)=-42\)
\(\Leftrightarrow5x^2+3x-5x-3=-42\)
\(\Leftrightarrow5x^2-2x+39=0\)
\(\Leftrightarrow25x^2-10x+195=0\)
\(\Leftrightarrow25x^2-10x+1=-194\)
\(\Leftrightarrow\left(5x-1\right)^2=-194\) (Khẳng định sai)
=> x không có giá trị.