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20 tháng 1 2019

a) \(x^3-3x^2+4=0\)

\(\Leftrightarrow\left(x-2\right)^2.\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

b) \(\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-5\right)-16=0\)

\(\Leftrightarrow4x^4-12x^3+7x^2+3x=0\)

\(\Leftrightarrow x\left(2x-3\right)\left(2x^2-3x-1\right)=0\)

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow2x=0+3\)

\(\Leftrightarrow2x=3\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

5 tháng 2 2018

a)  \(x^3-3x^2+4=0\)

\(\Leftrightarrow\)\(x^3+x^2-4x^2-4x+4x+4=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy....

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 \(x=3\pm\sqrt{5}\)

Bài 2:

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

  Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

  Vậy \(A_{Min}=-2\) khi \(x=-1\)

Bài 1:

a) Ta có: \(2x^2-6=0\)

\(\Leftrightarrow2x^2=6\)

\(\Leftrightarrow x^2=3\)

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

Vậy: \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)

10 tháng 12 2021

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)

13 tháng 8 2021

a)\(3x\left(x-1\right)+2x^2\left(x-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(3+2x\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\3+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=\dfrac{-3}{2}\end{matrix}\right.\)

a: Ta có: \(3x^2-3x+2x^3-2x^2=0\)

\(\Leftrightarrow2x^3+x^2-3x=0\)

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

\(\Leftrightarrow x\left(x-1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{3}{2}\end{matrix}\right.\)

b: Ta có: \(x^3+27=-x^2+9\)

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

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

\(\Leftrightarrow x+3=0\)

hay x=-3

18 tháng 2 2022

\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)

 

d: \(\Leftrightarrow3x^2-6x-2x+4=0\)

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

=>x=2 hoặc x=2/3

e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)

=>x(x-3)(x+1)=0

hay \(x\in\left\{0;3;-1\right\}\)

f: \(\Leftrightarrow x^2-5x-2+x=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow\left(x-2\right)^2=6\)

hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)

a. (3x - 1)2 - (x + 3)2 = 0

\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)

\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

 

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

11 tháng 10 2021

a: ta có: \(x^2+3x-\left(2x+6\right)=0\)

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

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

b: Ta có: \(5x+20-x^2-4x=0\)

\(\Leftrightarrow\left(x+4\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)

23 tháng 3 2023

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20 tháng 12 2019

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

27 tháng 10 2021

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)