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2(x - 3) + 5 = 3x - 1
2x-6+5=3x-1
2x-1=3x-1
2x-3x=-1+1
-x=0
x=0
2x(3x + 2) - 5 = 3( 2x^2 - 2x + 1)
6x2+4x-5=6x2-6x+3
6x2+4x-6x2+6x=3+5
10x=8
x=4/5
(3x - 2)(2x - 3) + 5 = 5
(3x-2)(2x-3)=0
=>3x-2=0 hoặc 2x-3=0
=>x=2/3 hoặc x=3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
1) 2x.(5x-3x)+2x.(3x-5)-3.(x-7)=3
10x-6x^2+6x^2-10x-3x+21=3
-3x =-18
suy ra x=6
2) 3x.(x+1) -2x.(x+2)=-1-x
3x^2 +3x-2x^2-4x =-1-x
x^2 =-1
suy ra không có giá trị nào của x thỏa mãn đề bài
3) 2x^2 +3.(x^2-1)=5x(x+1)
2x^2 +3x^2-3 =5x^2+5x
-5x =3
x=-3/5
giải rồi đấy
nhớ tích đúng nha :)
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a) \(\left|2x-5\right|=x+1\)
<=> \(\orbr{\begin{cases}2x-5=x+1\left(x\ge\frac{5}{2}\right)\\5-2x=x+1\left(x< \frac{5}{2}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\left(ktm\right)\\3x=4\end{cases}}\)
<=> \(x=\frac{4}{3}\left(tm\right)\)
b) \(\left|3x-2\right|-1=2x\) <=> \(\left|3x-2\right|=2x+1\)
<=> \(\orbr{\begin{cases}3x-2=2x+1\left(x\ge\frac{2}{3}\right)\\2-3x=2x+1\left(x< \frac{2}{3}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-3\left(ktm\right)\\5x=1\end{cases}}\) <=> \(x=\frac{1}{5}\left(tm\right)\)
c) \(\left|x-5\right|+5=x\) <=> \(\left|x-5\right|=x-5\)
<=> \(\orbr{\begin{cases}x-5=x-5\left(x\ge5\right)\\5-x=x-5\left(x< 5\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}0x=0\\2x=10\end{cases}}\) <=> 0x = 0 (luôn đúng) hoặc x = 5 (ktm)
Vậy x \(\ge\)5
d) \(\left|3x-5\right|=3x-5\) <=> \(\orbr{\begin{cases}3x-5=3x-5\left(x\ge\frac{5}{3}\right)\\5-3x=3x-5\left(x< \frac{5}{3}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}0x=0\left(luônđúng\right)\\6x=10\end{cases}}\)
<=> \(\orbr{\begin{cases}x\ge\frac{5}{3}\\x=\frac{5}{3}\left(ktm\right)\end{cases}}\)Vậy x \(\ge\)5/3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : |3x - 2| - 1 = x
=> |3x - 2| = x + 1
\(\Leftrightarrow\orbr{\begin{cases}3x-2=x+1\\3x-2=-x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-x=1+2\\3x+x=-1+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=3\\4x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{4}\end{cases}}\)
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\(a,\Rightarrow x\in\varnothing\left(\left|4+2x\right|\ge0>-4\right)\\ b,\Rightarrow\left|3x-1\right|=x-2\\ \Rightarrow\left[{}\begin{matrix}3x-1=x-2\left(x\ge\dfrac{1}{3}\right)\\3x-1=2-x\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Rightarrow x\in\varnothing\\ c,\Rightarrow\left|x+15\right|=3x-1\\ \Rightarrow\left[{}\begin{matrix}x+15=3x-1\left(x\ge-15\right)\\x+15=1-3x\left(x< -15\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=-\dfrac{7}{2}\left(ktm\right)\end{matrix}\right.\\ \Rightarrow x=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3x^2-3x\left(x-2\right)=36\\ \Leftrightarrow3x^2-3x^2+6x=36\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\\ b,5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x+2\right)=-36\\ \Leftrightarrow20x^3-10x^2+5x-20x^3+10x^2-4x+36=0\\ \Leftrightarrow\left(20x^3-20x^3\right)+\left(-10x^2+10x^2\right)+\left(5x-4x\right)=-36\\ \Leftrightarrow x=-36\)
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`@` `\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`
\(3x-|2x-1|=2\Rightarrow\orbr{\begin{cases}3x-\left(2x-1\right)=2\\3x-\left[\left(-2x\right)+1\right]=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-2x+1=2\\3x+2x-1=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x-2x=2-1\\3x+2x=2+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\5x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}}\)
Vậy \(x=1;\frac{3}{5}\)