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6 tháng 6 2015

x3-x2=4x2-8x+4

<=>x2(x-1)=4(x2-2x+1)

<=>x2(x-1)=4(x-1)2

<=>x2(x-1)-4(x-1)2=0

<=>(x-1)(x2-4x+4)=0

<=> (x-1)(x-2)2=0

<=>x-1=0 hoặc x-2=0

<=>x=1 hoặc x=2

x3 - x2 = 4x2 - 8x + 4

x3 - x2 - 4x2 + 8x - 4 = 0

x2(x - 1) - 4(x - 1)2 = 0

(x - 1)(x - 2)(x + 2) = 0

=> x - 1 = 0 hoặc x - 2 = 0 hoặc x + 2 = 0

=> x = 1 hoặc x = + 2

4 tháng 1 2017

Ta có: x3 – x2= x2(x -1); 4x2 – 8x + 4 = 4(x2 – 2x + 1) = 4(x – 1)2

Vậy x2 (x -1) = 4(x – 1)2 ⇒ x2(x -1) - 4(x – 1)2 = 0

⇒ (x – 1)(x2 – 4x + 4) = 0 ⇒ (x – 1)(x – 2)2 = 0

⇒ x – 1 = 0 hoặc x – 2 = 0 ⇒ x = 1 hoặc x = 2.

13 tháng 11 2021

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

18 tháng 8 2023

\(x^6+2x^3+1=0\)

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

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

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

\(\Leftrightarrow x=-1\)

___________

\(x\left(x-5\right)=4x-20\)

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

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

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

_____________

\(x^4-2x^2=8-4x^2\)

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

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

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

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

_______________

\(\left(x^3-x^2\right)-4x^2+8x-4\)

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

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

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

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

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

28 tháng 5

a: x(2x-7)+14=4x

=>x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(2x-7)(x-2)=0

=>\(\left[\begin{array}{l}2x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=2\end{array}\right.\)

b: \(25x^3=2x\)

=>\(25x^3-2x=0\)

=>\(x\left(25x^2-2\right)=0\)

TH1: x=0

=>x=0

TH2: \(25x^2-2=0\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

c: \(\left(x-5\right)^3=x^3-125\)

=>\(x^3-15x^2+75x-125=x^3-125\)

=>\(-15x^2+75x=0\)

=>-15x(x-5)=0

=>x(x-5)=0

=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)

d: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

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

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

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

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

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

28 tháng 5

a: x(2x-7)+14=4x

=>x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(2x-7)(x-2)=0

=>\(\left[\begin{array}{l}2x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=2\end{array}\right.\)

b: \(25x^3=2x\)

=>\(25x^3-2x=0\)

=>\(x\left(25x^2-2\right)=0\)

TH1: x=0

=>x=0

TH2: \(25x^2-2=0\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

c: \(\left(x-5\right)^3=x^3-125\)

=>\(x^3-15x^2+75x-125=x^3-125\)

=>\(-15x^2+75x=0\)

=>-15x(x-5)=0

=>x(x-5)=0

=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)

d: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

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

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

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

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

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

7 tháng 8 2021

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7 tháng 8 2021

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

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

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

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

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

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

23 tháng 10 2021

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

8 tháng 4 2022

b: 1/2x-4=0

=>1/2x=4

hay x=8

a: x+7=0

=>x=-7

e: 4x2-81=0

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

=>x=9/2 hoặc x=-9/2

g: x2-9x=0

=>x(x-9)=0

=>x=0 hoặc x=9

8 tháng 4 2022

a)\(x+7=0=>x=-7\)

b)\(\dfrac{1}{2}x-4=0=>\dfrac{1}{2}x=4=>x=8\)

c)\(-8x+20=0=>-8x=-20=>x=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100=>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)