Question

Tìm x

a. 2x(x-2)+x²=4

b. x³ +6x²+9x =0

c. 5x(x-2020)-x+2020=0

d. 4(x+1)² . (2x-1)²-8(x-1)(x+1)=11

e (x-3)(x+3x+9)+x(x+2)(2-x)=1

Answer 1

b) \(x^3+6x^2+9x=0\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=0\end{cases}}}\)

Vậy \(x\in\left\{-3;0\right\}\)

Answer 2

a) \(2x\left(x-2\right)+x^2=4\)

\(\Leftrightarrow2x\left(x-2\right)+x^2-4=0\)

\(\Leftrightarrow2x\left(x-2\right)+\left(x-2\right)\left(x+2\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)

Vậy \(x\in\left\{\frac{-2}{3};2\right\}\)

Answer 3

c) \(5x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2020=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{1}{5}\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{5};2020\right\}\)

Answer 4

Câu d. Em xem lại đề bài nhé 

d) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

<=> \(4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)

<=> \(4x^2+8x+4+4x^2-4x+1-8.x^2+8=11\)

<=> \(4x+13=11\)

<=> \(4x=-2\)

<=> \(x=-\frac{1}{2}\)

Vậy...

e) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)

<=> \(x^3-27+4x-x^3=1\)

<=> \(4x=28\Leftrightarrow x=7\)

Vậy...