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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 11 2023
|3\(x\) - 1| +|1 - 3\(x\)| = 9
vì |3\(x\) - 1| = |1 - 3\(x\)| nên:
|3\(x\) - 1| + |1 - 3\(x\)| = |3\(x\) - 1| + |3\(x\) - 1| = 2|3\(\)\(x\) - 1|
⇒2.|3\(x\) - 1| = 9
|3\(x\) - 1| = \(\dfrac{9}{2}\)
\(\left[{}\begin{matrix}3x-1=\dfrac{-9}{2}\\3x-1=\dfrac{9}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=-\dfrac{9}{2}+1\\3x=\dfrac{9}{2}+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=-\dfrac{7}{2}\\3x=\dfrac{11}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=\dfrac{11}{6}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- \(\dfrac{7}{6}\); \(\dfrac{11}{6}\)}
LM
Lê Minh Vũ
CTVHS
21 tháng 6 2023
\(2\left(3x-2\right)-3\left(x-2\right)=-1\)
\(6x-4-3x+6=-1\)
\(3x+2=-1\)
\(3x=-1-2\)
\(3x=-3\)
\(x=-1\)
\(2\left(3-3x^2\right):3x\left(2x-1\right)=9\)
\(6-6x^2:6x^2-3x=9\)
\(6-x^2-3x=9\)
\(-x^2-3x+6=9\)
\(-x^2-3x=5\)
\(-x\left(x+3\right)=5\)
\(x=-5;x=2\)
5 tháng 8 2021
\(\frac{7}{x-1}=\frac{9}{3x+1}\)
\(\Leftrightarrow\frac{7\left(3x+1\right)}{\left(x-1\right)\left(3x+1\right)}=\frac{9\left(x-1\right)}{\left(x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow7\left(3x+1\right)=9\left(x-1\right)\)
\(\Leftrightarrow21x+7=9x-9\)
\(\Leftrightarrow21x-9x=-9-7\)
\(\Leftrightarrow12x=\left(-16\right)\)
\(\Leftrightarrow x=\frac{-16}{12}\)
\(\Leftrightarrow x=\frac{-4}{3}\)
31 tháng 7 2023
\(4.3^x+3^{x+1}=63\)
\(\Rightarrow4.3^x+3.3^x=63\)
\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)
\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)
mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)
\(\Rightarrow x\in\varnothing\)
7 tháng 5 2023
`(3x-1)(x-3)-2(x-3)=9`
`-> 3x(x-3)-1(x-3)-2x+6=9`
`-> 3x^2-9x-x+3-2x+6=9`
`-> 3x^2-12x+9=9`
`-> 3x^2-12x=0`
`-> x(3x-12)=0`
`->`\(\left[{}\begin{matrix}x=0\\3x-12=0\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=0\\3x=12\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x={0 ; 4}`.
18 tháng 11 2016
a) |x - 5| - 2x = 3
| x - 5| = 3 + 2x
=> x - 5 = 3 + 2x hoặc x - 5 = -3 - 2x
=> -5 - 3 = 2x - x -5 + 3 = -2x - x
=> x = -8 -2 = -3x
=> x = 2/3
b) |2x - 1| + 3x = 1
|2x - 1| = 1 - 3x
=> 2x - 1 = 1 - 3x hoặc 2x - 1 = -1 + 3x
=> -1 - 1 = -3x - 2x -1 + 1 = 3x - 2x
=> -2 = -5x 0 = x
=> x = 2/5
c) | x - 5| = 3x - 2
=> x - 5 = 3x - 2 hoặc x - 5 = -3x + 2
=> -5 + 2 = 3x - x -5 - 2 = -3x - x
=> -3 = 2x -7 = -4x
=> x = -3/2 x = 7/4
d) |9 - 7x| = 5x - 3
=> 9 - 7x = 5x - 3 hoặc 9 - 7x = -5x + 3
=> 9 + 3 = 5x + 7x 9 - 3 = -5x + 7x
=> 12 = 12x 6 = 2x
=> x = 1 x = 3
16 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
30 tháng 12 2021
a x = \(\dfrac{-1}{12}\)
b x = \(\dfrac{-4}{3}\)
c x = \(\dfrac{-1}{6}\)
d x = \(\dfrac{-1}{4}\)
30 tháng 12 2021
\(\left(4x+1\right)^2=\dfrac{4}{9}\)
\(\left(4x+1\right)=\perp\left(\dfrac{2}{3}\right)^2\)
\(\text{Vậy }4x+1=\dfrac{2}{3}\)
\(4x\) \(=\dfrac{2}{3}+\left(-1\right)=\dfrac{-1}{3}\)
\(x\) \(=\left(\dfrac{-1}{3}\right).\dfrac{1}{4}=\dfrac{-1}{12}\)
\(\text{hoặc }4x+1=\dfrac{-2}{3}\)
\(4x\) \(=\left(\dfrac{-2}{3}\right)+\left(-1\right)=\dfrac{-5}{3}\)
\(x\) \(=\left(\dfrac{-5}{3}\right).\dfrac{1}{4}=\dfrac{-5}{12}\)
\(\Rightarrow x\in\left\{\dfrac{-1}{12};\dfrac{-5}{12}\right\}\)
\(\left(3x-1\right)^2=25\)
\(\left(3x-1\right)^2=\perp\left(5\right)^2\)
\(\text{Vậy }3x-1=5\)
\(3x\) \(=5+1=6\)
\(x\) \(=6:3=2\)
\(\text{hoặc }3x-1=-5\)
\(3x\) \(=\left(-5\right)+1=-4\)
\(x\) \(=\left(-4\right):3=\dfrac{-4}{3}\)
\(\Rightarrow x\in\left\{2;\dfrac{-4}{3}\right\}\)
\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\left(x-\dfrac{1}{3}\right)^2=\perp\left(\dfrac{1}{2}\right)^2\)
\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{2}\)
\(x\) \(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{2}\)
\(x\) \(=\left(\dfrac{-1}{2}\right)+\dfrac{1}{3}=\dfrac{-1}{6}\)
\(\Rightarrow x\in\left\{\dfrac{5}{6};\dfrac{-1}{6}\right\}\)
\(\left(4x-3\right)^2=16\)
\(\left(4x-3\right)=\perp\left(4\right)^2\)
\(\text{Vậy }4x-3=4\)
\(4x\) \(=4+3=7\)
\(x\) \(=7:4=\dfrac{7}{4}\)
\(\text{hoặc }4x-3=-4\)
\(4x\) \(=\left(-4\right)+3=-1\)
\(x\) \(=\left(-1\right):4=\dfrac{-1}{4}\)
\(\Rightarrow x\in\left\{\dfrac{7}{4};\dfrac{-1}{4}\right\}\)
\(9\left(3x+1\right)=\left(3x+1\right)\Leftrightarrow9\left(3x+1\right)-\left(3x+1\right)=0\)
\(\Leftrightarrow8\left(3x+1\right)=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)