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20 tháng 8 2019

b) \(x^2-\left(x+4\right)\left(x+3\right)=24\)

\(\Leftrightarrow x^2-\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow x^2-x^2-7x-12=24\)

\(\Leftrightarrow-7x-12=24\)

\(\Leftrightarrow-7x=36\)

\(\Leftrightarrow x=\frac{-36}{7}\)

20 tháng 8 2019

a) \(x^2-x\left(2x+3\right)=2x-x^2+1\)

\(\Leftrightarrow x^2-2x^2-3x=2x-x^2+1\)

\(\Leftrightarrow5x+1=0\)

\(\Leftrightarrow x=\frac{-1}{5}\)

27 tháng 2 2021

`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`

a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)

23 tháng 9 2021

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

23 tháng 9 2021

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

16 tháng 2 2022

\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (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: |2x-3|=1

=>2x-3=1 hoặc 2x-3=-1

=>x=1(nhận) hoặc x=2(loại)

KHi x=1 thì \(A=\dfrac{1+1^2}{2-1}=2\)

b: ĐKXĐ: x<>-1; x<>2

\(B=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x-2\right)\left(x+1\right)}=\dfrac{-x+2}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{x+1}\)

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)

a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)

b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)

c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)

\(P-2=\dfrac{2x-2x-2}{x+1}=\dfrac{-2}{x+1}\)

P<=2

=>x+1>0

=>x>-1

a: Ta có: \(N=\dfrac{x^3-1}{x^2-2x+1}\)

\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)^2}\)

\(=\dfrac{x^2+x+1}{x-1}\)

\(=\dfrac{\left(-1\right)^2+\left(-1\right)+1}{-1-1}=\dfrac{1}{-2}=-\dfrac{1}{2}\)

b: Ta có: \(M=\dfrac{x^3+8}{x^2-2x+4}\)

\(=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}\)

\(=x+2=0\)

25 tháng 8 2021

a) \(N=\dfrac{x^3-1}{x^2-2x+1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)^2}=\dfrac{x^2+x+1}{x-1}=\dfrac{\left(-1\right)^2-1+1}{-1-1}=-\dfrac{1}{2}\)b) \(M=\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2=-2+2=0\)

a: 3x^3+2x^2-7x+a chia hêt cho 3x-1

=>3x^3-x^2+3x^2-x-6x+2+a-2 chia hết cho 3x-1

=>a-2=0

=>a=2

c: =>2x^2-6x+(a+6)x-3a-18+3a+19 chia x-3 dư 4

=>3a+19=4

=>3a=-15

=>a=-5

d: 2x^3-x^2+ax+b chiahêt cho x^2-1

=>2x^3-2x-x^2+1+(a+2)x+b-1 chia hết cho x^2-1

=>a+2=0 và b-1=0

=>a=-2 và b=1