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a. \(5.\left(x-2\right)+3.\left(x-2\right)=0\)
\(\Rightarrow8.\left(x-2\right)=0\)
\(\Rightarrow x-2=0:8\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy...
b. \(\dfrac{2}{3}+\dfrac{5}{2}:x=\dfrac{2}{4}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{2}{4}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{-1}{6}\)
\(\Rightarrow x=\dfrac{5}{2}:\dfrac{-1}{6}=-15\)
Vậy...
c. \(2.\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow x-\dfrac{1}{7}=0:2\)
\(\Rightarrow x-\dfrac{1}{7}=0\)
\(\Rightarrow x=\dfrac{1}{7}\)
Vậy...
d. \(\dfrac{11}{20}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}:\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{-3}{20}\)
Vậy...
e. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)
Vậy...
g. \(\dfrac{2}{3}x+\dfrac{5}{7}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{-29}{70}\)
\(\Rightarrow x=\dfrac{-29}{70}:\dfrac{2}{3}=\dfrac{-87}{140}\)
Vậy...
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
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
a) \(\frac{3}{4}-\left(\frac{1}{2}:x+\frac{1}{2}\right)=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{3}{4}-\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{15}{20}-\frac{12}{20}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{13}{20}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{10}{20}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{20}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{3}{20}\)
\(\Leftrightarrow x=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\)
Vậy: \(x=\frac{10}{3}\)
b) \(3x.\left(\frac{1}{2}.x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\\frac{1}{2}x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
Vậy: \(x\in\left\{0;2\right\}\)
c) \(\left(4-x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\2x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=\frac{3}{2}\end{cases}}}\)
Vậy: \(x\in\left\{4;\frac{3}{2}\right\}\)
d) \(\frac{4}{-3}=\frac{-12}{x}\)
\(\Leftrightarrow4x=\left(-12\right).\left(-3\right)\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
Vậy: \(x=9\)
e) \(\frac{4x}{-3}=\frac{12}{-x}\)
\(\Leftrightarrow4x.\left(-x\right)=12.\left(-3\right)\)
\(\Leftrightarrow-4x^2=-36\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy: \(x\in\left\{3;-3\right\}\)
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy....
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
a.Ta có: 2|x| + 3|1-x| - |x-3| =0
2x + 3 - 3x - x +3 =0
-2x + 6 =0
-2x =-6
x =3
b.Ta có:4|3x-1| + |x| - 2|x-5| +7|x-3| =12
12x-4 + x -2(x-5) + 7x - 3 =12
12x-4+ x - 2x +10 + 7x - 3=12
18x + 3 =12
18x =12-3
18x =9
x =1/2
a. 2 |x |+ 3 | 1- x| - 5 | x - 3= 0
2 x + 3 - 3 x - x + 3 = 0
-2x + 6 = 0
- 2 x = - 6
x = 3