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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
Trả lời :
a, \(\frac{3}{4}-\left(\frac{1}{2}\div x+\frac{1}{2}\right)=\frac{3}{5}\)
=> \(\frac{1}{2}\div x+\frac{1}{2}=\frac{3}{20}\)
=> \(\frac{1}{2}\div x=\frac{-7}{20}\)
=> \(x=\frac{-10}{7}\)
b, (4 - x) . (2x + 3) = 0
=> \(\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=\frac{-3}{2}\end{cases}}\)
c, \(\frac{4}{-3}=\frac{-12}{x}\)
=> 4x = 36
=> x = 9
d, \(\frac{4x}{-3}=\frac{12}{-x}\)
=> \(-4x^2=-36\)
=> 4x2 = 36
=> x2 = 9
=> x = \(\pm3\)
|\(x-\dfrac{1}{2}\)| + 2\(x\) = 6
|\(x-\dfrac{1}{2}\)| = 6 - 2\(x\); 6 - 2\(x\) > 0 ⇒ 6 > 2\(x\) ⇒ \(x\) < 3
\(\left[{}\begin{matrix}x-\dfrac{1}{2}=6-2x\\x-\dfrac{1}{2}=-6+2x\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+2x=6+\dfrac{1}{2}\\2x-x=6-\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=\dfrac{13}{2}\\x=\dfrac{11}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{11}{2}\end{matrix}\right.\)
\(x=\dfrac{11}{2}\) > 3 (loại)
Vậy \(x\) = \(\dfrac{13}{6}\)
a, |x^2 - 3x| = 0
=> x^2 - 3x = 0
=> x(x - 3) = 0
=> x = 0 hoặc x - 3 = 0
=> x = 0 hoặc x = 3
vậy_
\(\left|a^2-3a\right|=0\)
\(\Rightarrow a^2-3a=0\)
\(\Rightarrow a\left(a-3\right)=0\)
\(\Rightarrow\hept{\begin{cases}a=0\\a=3\end{cases}}\)
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...
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,Ta có: 3/4-(1/2:x+1/2)=3/5
-(1/2:x+1/2)=3/5-3/4
-(1/2:x+1/2)=-3/20
1/2:x+1/2=3/20
1/2:x=3/20-1/2
1/2:x=-7/20
x=1/2:-7/20
x=-10/7
b,Ta có: 3x.(1/2x-1)=0
Với 3x=0 =>x=0
vói1/2x-1=0