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a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
Lời giải:
a.
a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$
$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$
$\Leftrightarrow 3x+1=5x-3$
$\Leftrightarrow 4=2x$
$\Leftrightarrow x=2$
b.
$(2x-1)(x+3)-(x-2)(x+3)=3x+1$
$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$
$\Leftrightarrow x^2+5x+1=3x+1$
$\Leftrightarrow x^2+2x=0$
$\Leftrightarrow x(x+2)=0$
$\Leftrightarrow x=0$ hoặc $x=-2$
c.
$x^2(x-1)-x(x-1)(x+1)=0$
$\Leftrightarrow x^2(x-1)-(x^2+x)(x-1)=0$
$\Leftrightarrow (x-1)[x^2-(x^2+x)]=0$
$\Leftrightarrow (x-1)(-x)=0$
$\Leftrightarrow x-1=0$ hoặc $-x=0$
$\Leftrightarrow x=1$ hoặc $x=0$
d.
$4x(x-5)-(2x-3)(2x+3)=9$
$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$
$\Leftrightarrow -20x=0$
$\Leftrightarrow x=0$
a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x+3x+3-5x+3=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow2x=4\)
hay x=2
b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)
\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)
\(\Leftrightarrow4x^2-20x-4x^2+9=9\)
hay x=0
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).
\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)
\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)
\(d)2x\left(x-2\right)+5\left(x-2\right)=0.\\ \Leftrightarrow\left(2x+5\right)\left(x-2\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}.\\x=2.\end{matrix}\right.\)
\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)
\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)
\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)
\(\Leftrightarrow x=1\left(koTM\right).\)
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
b)x+3=4:2
=> x=-1
d)5x-15=3x-5
<=> 5x-3x=15-5
<=> 2x=10
<=> x=5
f) 35-7x=11-5x
<=> 35-11=-5x+7x
<=> 24=2x
<=> x=12
h) 6x-2-3x=10
<=> 3x=10+2
<=> x=4
j)3-2x=3x+3-x-2
<=> 3-2x=2x+1
<=>-4x=-2
<=> x=1/2
\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)
a. (x + 2)2 + (x + 3)2 - 2(x - 2)(x - 3) = 19
<=> (x2 + 4x + 4) + (x2 + 6x + 9) - (2x + 4)(x - 3) = 19
<=> x2 + 4x + 4 + x2 + 6x + 9 - 2x2 + 6x - 4x + 12 = 19
<=> x2 + x2 - 2x2 + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0
<=> 12x + 6 = 0
<=> 6(2x + 1) = 0
<=> 2x + 1 = 0
<=> 2x = -1
<=> x = \(\dfrac{-1}{2}\)
`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`