Bài 4: Phương trình tích

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)

-Vậy \(S=\left\{4;1;-1\right\}\)

`4x^2 + ( x - 1 )^2 - ( 2x + 1 )^2 = 0`

`<=> 4x^2 + x^2 - 2x + 1 - 4x^2 - 4x - 1 = 0`

`<=> x^2 - 6x = 0`

`<=> x ( x - 6 ) = 0`

`<=>` $\left[\begin{matrix} x= 0\\ x - 6 =0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x= 0\\ x =6\end{matrix}\right.$

Vậy `S = { 0 ; 6 }`

\(\Leftrightarrow4x^2+\left(x^2-2x+1\right)-\left(4x^2+4x+1\right)=0\)

\(\Leftrightarrow4x^2+x^2+2x-1-4x^2-4x-1=0\)

\(\Leftrightarrow x^2-2x-2=0\) (ko tính đc)

Bạn cần viết đề bằng công thức toán ( biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.

\(x=4\)

\(2\left(2m-3\right)\left(m+1\right)\sqrt{4}=\dfrac{3}{4}-\dfrac{m}{2}\)

\(\Leftrightarrow2\left(2m^3+2m-3m-3\right).2=\dfrac{3-2m}{4}\)

\(\Leftrightarrow8m^2+8m-12m-12=\dfrac{3-2m}{4}\)

\(\Leftrightarrow32m^2-16m-48=3-2m\)

\(\Leftrightarrow32m^2-14m-51=0\)

\(\Delta=\left(-14\right)^2-4.32-\left(-51\right)\)

   \(=196+6528=6724>0\)

\(\rightarrow\left\{{}\begin{matrix}x_1=\dfrac{3}{2}\\x_2=-\dfrac{17}{16}\end{matrix}\right.\)

đây là kiến thức có trong bài thi đấy ạ !

 giải Pt: x^2 - 4 + 3 (x-2) = 0

<=> x^2-4 + 3x-6 =0 

<=> x^2+3x = 6+4

<=> 4x = 10 

         x = 5/2

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

\(3x^2+y^2+2x-2y-1=0\)

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

\(\Leftrightarrow x^2+2-2xy+y^2+2x-2y-1=0\)

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

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

\(\Leftrightarrow x-y+1=0\)

\(\Leftrightarrow y=x+1\)

Thế vào \(x\left(x+y\right)=1\)

\(\Rightarrow x\left(2x+1\right)=1\)

\(\Leftrightarrow2x^2+x-1=0\Rightarrow\left[{}\begin{matrix}x=-1\Rightarrow y=0\\x=\dfrac{1}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)

Lời giải:
$x^2-4+(x-2)(x+3)=0$

$\Leftrightarrow (x-2)(x+2)+(x-2)(x+3)=0$
$\Leftrightarrow (x-2)(x+2+x+3)=0$

$\Leftrightarrow (x-2)(2x+5)=0$

$\Rightarrow x-2=0$ hoặc $2x+5=0$

$\Leftrightarrow x=2$ hoặc $x=-\frac{5}{2}$

\(=\left(x^2+x+4\right)^2+2\cdot\left(x^2+x+4\right)\cdot4x+\left(4x\right)^2\)

\(=\left(x^2+5x+4\right)^2\)

\(=\left(x+1\right)^2\cdot\left(x+4\right)^2\)