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\(x^3+31x^2+155x+125=0\)

\(\Leftrightarrow x^3+3.x^2.5+3.x.5^2+125+16x^2+80x=0\)

\(\Leftrightarrow\left(x+5\right)^3+16x\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right).\left[\left(x+5\right)^2+16x\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\\left(x+5\right)^2+16x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x^2+26x+25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x\left(x+26\right)=-25\left(2\right)\end{matrix}\right.\)

Đến đây bạn giải tiếp hệ (2) sẽ ra được x.

29 tháng 11 2021

ảnh lỗi r ạ

29 tháng 11 2021

Lỗi rùi

10 tháng 4

8: \(x^3-4x^2+8x-32=0\)

=>\(x^2\left(x-4\right)+8\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2+8\right)=0\)

=>x-4=0

=>x=4

9: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

=>\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

=>\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

=>\(\left(x+3\right)\left(x^2-2x\right)=0\)

=>x(x-2)(x+3)=0

=>x∈{0;2;-3}

10: \(x^2-10x+16=0\)

=>\(x^2-2x-8x+16=0\)

=>x(x-2)-8(x-2)=0

=>(x-2)(x-8)=0

=>x=2 hoặc x=8

11: \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

=>\(2x^2+3\left(x^2-1\right)=5x^2+5x\)

=>\(5x^2+5x=2x^2+3x^2-3\)

=>5x=-3

=>\(x=-\frac35\)

12: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=0\)

=>\(x^2-9-\left(x^2+3x-10\right)=0\)

=>\(x^2-9-x^2-3x+10=0\)

=>-3x+1=0

=>-3x=-1

=>\(x=\frac13\)

18 tháng 8 2023

a: =>2x^2+9x-6x-27=0

=>x(2x+9)-3(2x+9)=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

b: =>-10x^2+6x-5x+3=0

=>-2x(5x-3)-(5x-3)=0

=>(5x-3)(-2x-1)=0

=>x=-1/2 hoặc x=5/3

c: =>-x^3+2x^2-x^2+4=0

=>-x^2(x-2)-(x-2)(x+2)=0

=>(x-2)(-x^2-x-2)=0

=>x-2=0

=>x=2

d: =>(x^3+8)-4x(x+2)=0

=>(x+2)(x^2-2x+4)-4x(x+2)=0

=>(x+2)(x^2-6x+4)=0

=>x=-2 hoặc \(x=3\pm\sqrt{5}\)

29 tháng 10 2021

g) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

  \(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

  \(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

  \(\Rightarrow-2\left(2x-5\right)=0\Rightarrow x=\dfrac{5}{2}\)

i) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

  \(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

  \(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

  \(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\Rightarrow x\left(x+3\right)\left(x-2\right)=0\)

  \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

19 tháng 12 2021

a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

23 tháng 10 2021

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

`P(x)=\(4x^2+x^3-2x+3-x-x^3+3x-2x^2\)

`= (x^3-x^3)+(4x^2-2x^2)+(-2x-x+3x)+3`

`= 2x^2+3`

 

`Q(x)=`\(3x^2-3x+2-x^3+2x-x^2\)

`= -x^3+(3x^2-x^2)+(-3x+2x)+2`

`= -x^3+2x^2-x+2`

`P(x)-Q(x)-R(x)=0`

`-> P(X)-Q(x)=R(x)`

`-> R(x)=P(x)-Q(x)`

`-> R(x)=(2x^2+3)-(-x^3+2x^2-x+2)`

`-> R(x)=2x^2+3+x^3-2x^2+x-2`

`= x^3+(2x^2-2x^2)+x+(3-2)`

`= x^3+x+1`

`@`\(\text{dn inactive.}\)

18 tháng 4 2023

a: P(x)-Q(x)-R(x)=0

=>R(x)=P(x)-Q(x)

=2x^2+3+x^3-2x^2+x-2

=x^3+x+1