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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 \(x=-\dfrac{7}{30}\)
\(1,\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow3x-6=4x+4\)
\(\Rightarrow3x-4x=4+6\)
\(\Rightarrow-x=10\Leftrightarrow x=-10\)
\(2,\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\Leftrightarrow x=7\)
\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow72x-64x=24+96\)
\(\Rightarrow8x=120\)
\(\Rightarrow x=15\)
A) 5/4+x=2/3
B) -x-2=5/4
C)4x+1/3=3/2
Đ) 1/3-2/5+3x=3/4
E) 3x+7+2x=4x-3
G) 3x(2x-3)-2x(3x-4)=15
H) x^2-x=0
a) \(x=-\frac{7}{12}\)
b) \(x=-\frac{13}{4}\)
c) \(x=\frac{7}{24}\)
d) \(x=\frac{49}{180}\)
e) \(x=-10\)
g) \(x=15\)
h) \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
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 \(S=\left\{-3;-2\right\}\)
* 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) \(\frac{25}{12}.x+\frac{11}{15}=\frac{9}{10}\)
=> \(\frac{25}{12}.x=\frac{9}{10}-\frac{11}{15}\)
=> \(\frac{25}{12}.x=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\frac{25}{12}\)
=> \(x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\).
3) \(\frac{29}{12}.\left[x\right]-\frac{5}{6}=\frac{3}{8}\)
=> \(\frac{29}{12}.\left[x\right]=\frac{3}{8}+\frac{5}{6}\)
=> \(\frac{29}{12}.x=\frac{29}{24}\)
=> \(x=\frac{29}{24}:\frac{29}{12}\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\).
4) \(\left[4x+\frac{3}{4}\right]-\frac{5}{4}=2\)
=> \(\left[4x+\frac{3}{4}\right]=2+\frac{5}{4}\)
=> \(4x+\frac{3}{4}=\frac{13}{4}\)
=> \(4x=\frac{13}{4}-\frac{3}{4}\)
=> \(4x=\frac{5}{2}\)
=> \(x=\frac{5}{2}:4\)
=> \(x=\frac{5}{8}\)
Vậy \(x=\frac{5}{8}\).
5) 2x + 2x+3 = 144
⇔ 2x + 2x . 23 = 144
⇔ 2x . (1 + 23) = 144
⇔ 2x . 9 = 144
⇔ 2x = 144 : 9
⇔ 2x = 16
⇔ 2x = 24
=> x = 4
Vậy x = 4.
Chúc bạn học tốt!
1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow-2x\left(x+1\right)=0\)
Vì -2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;-1}
2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)
\(\Leftrightarrow11x-2=0\)
\(\Leftrightarrow11x=2\)
hay \(x=\frac{2}{11}\)
Vậy: \(x=\frac{2}{11}\)
3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)
\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)
hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)
Vậy: \(x=\frac{-13}{4}\)
4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)
1. \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x=3x^2-x\)
\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)
\(\Leftrightarrow-2x-2x^2=0\)
\(\Leftrightarrow-2x\left(1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2. \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x=12-x\)
\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)
\(\Leftrightarrow-2+11x=0\)
\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)
3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)
\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)
4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)
\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)
\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)