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\(a,\frac{3x+2}{5x+7}=\frac{3x-1}{5x-1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x-1\right)}=\frac{3}{8};\frac{3x+2}{5x+7}=\frac{3}{8}\Leftrightarrow24x+16=15x+21\Leftrightarrow9x=5\Leftrightarrow x=\frac{5}{9}\) \(b,\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow37.7-7x=3x+39\Leftrightarrow259-7x=3x+39\Leftrightarrow220-7x=3x\Leftrightarrow10x=220\Leftrightarrow x=22\) \(c,\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{x+4}{2x+6}=\frac{\left(x+4\right)-\left(x+1\right)}{2x+6-\left(2x+1\right)}=\frac{3}{5};\frac{x+1}{2x+1}=\frac{3}{5}\Leftrightarrow5x+5=6x+3\Leftrightarrow x=2\) \(d,\frac{x-2}{x+2}=\frac{x+3}{x-4}=\frac{\left(x+3\right)-\left(x-2\right)}{\left(x-4\right)-\left(x+2\right)}=\frac{5}{-6};\frac{x-2}{x+2}=\frac{5}{-6}\Leftrightarrow6\left(2-x\right)=5x+10\Leftrightarrow2-6x=5x\Leftrightarrow x=\frac{2}{11}\) \(f,\frac{3x-5}{x}=\frac{9x}{3x+2}=\frac{9x-15}{3x}=\frac{9x-\left(9x-15\right)}{\left(3x+2\right)-3x}=\frac{15}{2};\frac{9x}{3x+2}=\frac{15}{2}\Leftrightarrow18x=45x+30\Leftrightarrow27x+30=0\Leftrightarrow x=\frac{-10}{9}\) \(e,\frac{x+2}{6}=\frac{5x-1}{5}\Leftrightarrow5\left(x+2\right)=6\left(5x-1\right)\Leftrightarrow5x+10=30x-6\Leftrightarrow10=25x-6\Leftrightarrow25x=16\Leftrightarrow x=\frac{16}{25}\)
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)
=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)
=> \(\left|2-\frac{3}{2}x\right|=x+6\)
ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)
Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)
=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)
=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)
b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)
=> x = 1/4
hoặc x = 0 hoặc x = 1/2
\(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Rightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Rightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Rightarrow x=-\frac{87}{140}\)
tíc mình nha
a)\(\frac{3x+2}{7}=\frac{4x-5}{6}\)
\(\Leftrightarrow\frac{6\left(3x+2\right)}{42}=\frac{7\left(4x-5\right)}{42}\)
\(\Leftrightarrow6\left(3x+2\right)=7\left(4x-5\right)\)
\(\Leftrightarrow18x+12=28x-35\)
\(\Leftrightarrow18x-28x=-12-35\)
\(\Leftrightarrow-10x=-47\Leftrightarrow x=\frac{47}{10}\)
b) \(\frac{x+1}{x+3}=\frac{x+2}{x+4}\left(đkxđ:x\ne-3,-4\right)\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=\left(x+2\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)-\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+4x+x+4-x^2-3x+2x+6=0\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left(4x+4\right)+\left(4+6\right)+\left(x-3x\right)=0\)
\(\Leftrightarrow4x+4+10-2x=0\)
\(\Leftrightarrow4x+14-2x=0\)
\(\Leftrightarrow2x=-14\Leftrightarrow x=-7\)
a) \(\frac{3x+2}{7}=\frac{4x-5}{6}\)
=> \(6\left(3x+2\right)=7\left(4x-5\right)\)
=> \(18x+12=28x-35\)
=> \(18x+12-28x+35=0\)
=> \(\left(18x-28x\right)+\left(12+35\right)=0\)
=> \(-10x+47=0\)
=> \(-10x=-47\Rightarrow x=\frac{47}{10}\)
b) \(\frac{x+1}{x+3}=\frac{x+2}{x+4}\)
=> (x + 1)(x + 4) = (x + 2)(x + 3)
=> x(x + 4) + 1(x + 4) = x(x + 3) + 2(x + 3)
=> x2 + 4x + x + 4 = x2 + 3x + 2x + 6
=> x2 + 5x + 4 = x2 + 5x +6
=> x2 + 5x + 4 - x2 - 5x - 6 = 0
=> (x2 - x2) + (5x - 5x) + (4 - 6) = 0
=> -2 \(\ne\)0
=> không tìm được x thỏa mãn
hay cách khác : \(\frac{x+1}{x+3}=\frac{x+2}{x+4}\)
=> \(\frac{x+3-2}{x+3}=\frac{x+4-2}{x+4}\)
=> \(1-\frac{2}{x+3}=1-\frac{2}{x+4}\)
=> \(1-\frac{2}{x+3}-\left(1-\frac{2}{x+4}\right)=0\)
=> \(1-\frac{2}{x+3}-1+\frac{2}{x+4}=0\)
=> \(\left(1-1\right)+\left(-\frac{2}{x+3}+\frac{2}{x+4}\right)=0\)
=> \(-\frac{2}{x+3}+\frac{2}{x+4}=0\)
=> \(\frac{-2\left(x+4\right)+2\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}=0\)
=> \(-2x-8+2x+6=0\)
=> \(\left(-2x+2x\right)+\left(-8+6\right)=0\)
=> \(-2\ne0\)
=> không tìm được x thỏa mãn