(4x-8)(3x-6) > (4x-8)(2x+2)
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`2//(5x-8)-3(4x-5)=4(3x-4)`
`<=>5x-8-12x+15=12x-16`
`<=>-19x=-23`
`<=>x=23/19` Vậy `x=23/19`
`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`
`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`
`<=>4x=8`
`<=>x=2` Vậy `x=2`
a) \(4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\4x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\4x=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{4}=2\end{matrix}\right.\)
b) \(x^2+2x-3x-6=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
a: (x+1)(4x-8)=0
=>x+1=0 hoặc 4x-8=0
=>x=2 hoặc x=-1
b: =>x(x+2)-3(x+2)=0
=>(x+2)(x-3)=0
=>x=3 hoặc x=-2
a )
\(5x\left(4x-5\right)-4x\left(5x-6\right)=30\)
\(\Rightarrow20x^2-25x-20x^2+24x=30\)
\(\Rightarrow-x=30\)
\(\Rightarrow x=-30\)
Vậy ...
b )
\(2x\left(6-3x\right)+3x\left(2x-5\right)=12\)
\(\Rightarrow12x-6x^2+6x^2-15x=12\)
\(\Rightarrow-3x=12\)
\(\Rightarrow x=-4\)
Vậy ...
a) \(5x\left(4x-5\right)-4x\left(5x-6\right)-30\)
\(\Rightarrow20x^2-25x-20x^2+24x=30\)
\(\Rightarrow-1x=30\)
\(\Rightarrow x=-30\)
Vậy x = -30
b) \(2x\left(6-3x\right)+3x\left(2x-5\right)=12\)
\(\Rightarrow12x-6x^2+6x^2-15x=12\)
\(\Rightarrow-3x=12\)
\(\Rightarrow x=-4\)
Vậy x = -4
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
Lời giải:
1.
$4x+9=0$
$4x=-9$
$x=\frac{-9}{4}$
2.
$-5x+6=0$
$-5x=-6$
$x=\frac{6}{5}$
3.
$x^2-1=0$
$x^2=1=1^2=(-1)^2$
$x=\pm 1$
4.
$x^2-9=0$
$x^2=9=3^2=(-3)^2$
$x=\pm 3$
5.
$x^2-x=0$
$x(x-1)=0$
$x=0$ hoặc $x-1=0$
$x=0$ hoặc $x=1$
6.
$x^2-2x=0$
$x(x-2)=0$
$x=0$ hoặc $x-2=0$
$x=0$ hoặc $x=2$
7.
$x^2-3x=0$
$x(x-3)=0$
$x=0$ hoặc $x-3=0$
$x=0$ hoặc $x=3$
8.
$3x^2-4x=0$
$x(3x-4)=0$
$x=0$ hoặc $3x-4=0$
$x=0$ hoặc $x=\frac{4}{3}$
\(\left(4x-8\right)\left(3x-6\right)>\left(4x-8\right)\left(2x+2\right)\)
\(\Leftrightarrow\)\(\left(4x-8\right)\left(3x-6\right)-\left(4x-8\right)\left(2x+2\right)>0\)
\(\Leftrightarrow\)\(\left(4x-8\right)\left(3x-6-2x-2\right)>0\)
\(\Leftrightarrow\)\(4\left(x-2\right)\left(x-8\right)>0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x-2>0\\x-8>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2< 0\\x-8< 0\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x>2\\x>8\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2\\x< 8\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x>8\\x< 2\end{cases}}\)
Vậy...
8 < x < 2 sao được ạ?