Tìm x biết: \(|2x+3|-4x< 9\)
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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`
\(4x^2-9+\left(2x-3\right)\left(x+7\right)=0\)
\(\Rightarrow\left(4x^2-9\right)+\left(2x-3\right)\left(x+7\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(2x+3\right)+\left(2x-3\right)\left(x+7\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(2x+3+x+7\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x+10\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\3x+10=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
\(4x^2-9=3x\left(2x-3\right)\)
\(\Rightarrow\left(2x-3\right)\left(2x+3\right)-3x\left(2x-3\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(-x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
\(\Rightarrow\left(2x-3\right)\left(2x+3\right)-3x\left(2x-3\right)=0\\ \Rightarrow\left(2x-3\right)\left(2x+3-3x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
\(\left(2x-3\right)^2-9.\left(4x^2-9\right)^2=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(3.\left(4x^2-9\right)\right)^2=0\)
\(\Leftrightarrow\left(2x-3-3.\left(4x^2-9\right)\right).\left(2x-3+3.\left(4x^2-9\right)\right)=0\)
\(\Leftrightarrow\left(2x-3-3.\left(2x-3\right).\left(2x+3\right)\right).\left(2x-3+3.\left(2x-3\right).\left(2x+3\right)\right)=0\)
\(\Leftrightarrow\left(\left(2x-3\right).\left(1-3.\left(2x+3\right)\right)\right).\left(\left(2x-3\right).\left(1+3.\left(2x+3\right)\right)\right)=0\)
\(\Leftrightarrow\left(\left(2x-3\right).\left(-8-6x\right)\right).\left(\left(2x-3\right).\left(10+6x\right)\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2.\left(-8-6x\right).\left(10+6x\right)=0\)
Tiếp tục giải ra thôi, dễ quá đi mất
\(\)
\(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x\right)^2-3^2-x\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)-x\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left[\left(2x+3\right)-x\right]=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)
4x^2-9-x.(2x-3)=0
<=>4x2-9-2x2+3x=0
<=>2x2+3x-9=0
<=>2x2-3x+6x-9=0
<=>x.(2x-3)+3.(2x-3)=0
<=>(2x-3)(x+3)=0
<=>2x-3=0 hoặc x+3=0
<=>x=3/2 hoặc x=-3
=>\(\orbr{\begin{cases}2x+3-4x< 9\\-\left(2x+3\right)-4x< 9\end{cases}}\)
=>\(\orbr{\begin{cases}6x< 6\\-2x-3-4x< 9\end{cases}}\)
=>\(\orbr{\begin{cases}x< 1\\-6x< 12\end{cases}}\)
=>\(\orbr{\begin{cases}x< 1\\x< -2\end{cases}}\)
=>x<-2
Vậy chỉ cần x <-2 là được