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a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
1, \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=5\)
\(\Leftrightarrow4x^2+12x+9-4x^2-1=5\)
\(\Leftrightarrow12x=-3\)
\(\Leftrightarrow x=\dfrac{-1}{4}\)
Vậy \(x=\dfrac{-1}{4}\)
2, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\)
\(\Leftrightarrow x^3+27-x^3-5x=20\)
\(\Leftrightarrow5x=7\)
\(\Leftrightarrow x=\dfrac{7}{5}\)
Vậy...
5, \(x^2-9+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
Vậy...
1) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=5\) (1)
\(\Leftrightarrow4x^2+12x+9-\left(4x^2-1\right)=5\)
\(\Leftrightarrow4x^2+12x+9-4x^2+1=5\)
\(\Leftrightarrow12x+10=5\)
\(\Leftrightarrow12x=5-10\)
\(\Leftrightarrow12x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{5}{12}\right\}\)
2) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\) (2)
\(\Leftrightarrow x^3+27-x^3-5x=20\)
\(\Leftrightarrow27-5x=20\)
\(\Leftrightarrow-5x=20-27\)
\(\Leftrightarrow-5x=-7\)
\(\Leftrightarrow x=\dfrac{7}{5}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{7}{5}\right\}\)
3) \(\left(x+2\right)^3-x\left(x^2+6x\right)=15\) (3)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=15\)
\(\Leftrightarrow12x+8=15\)
\(\Leftrightarrow12x=15-8\)
\(\Leftrightarrow12x=7\)
\(\Leftrightarrow x=\dfrac{7}{12}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{7}{12}\right\}\)
4) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+10\right)\left(x-1\right)=7\) (4)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x\left(x+10\right)\right)=7\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2-10x\right)=7\)
\(\Leftrightarrow\left(x-1\right)\left(-9x+1\right)=7\)
\(\Leftrightarrow-9x^2+x+9x-1=7\)
\(\Leftrightarrow-9x^2+10-1=7\)
\(\Leftrightarrow-9x^2+10x-1-7=0\)
\(\Leftrightarrow-9x^2+10x-8=0\)
\(\Leftrightarrow9x^2-10x+8=0\)
\(\Leftrightarrow x\notin R\)
5) \(x^2-9+5\left(x+3\right)=0\) (5)
\(\Leftrightarrow x^2-9+5x+15=0\)
\(\Leftrightarrow x^2+5x+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+1}{2}\\x=\dfrac{-5-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình (5) là \(S=\left\{-3;-2\right\}\)