Question

[Xuyen Phan]

Giải giúp mình với ạ

Đề: giải các pt bậc hai 1 ẩn sau  

Answer 1

[Di Di]

`x^2 -x=12`

`<=>x^2 -x-12=0`

`<=> x^2+3x-4x-12=0`

`<=> x(x+3)-4(x+3)=0`

`<=>(x+3)(x-4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

`---`

`2x^2-3x=15-4x`

`<=> 2x^2-3x+4x=15`

`<=>2x^2 +x-15=0`

`<=>2x^2+6x-5x-15=0`

`<=> 2x(x+3)-5(x+3)=0`

`<=>(x+3)(2x-5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

`---`

`x(x-5)=24`

`<=> x^2 -5x-24=0`

`<=>x^2+3x-8x-24=0`

`<=>x(x+3) -8(x+3)=0`

`<=>(x+3)(x-8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)

`----`

`x(x-3)=10(x-4)`

`<=> x^2 -3x =10x -40`

`<=>x^2 -3x-10x +40=0`

`<=> x^2 -13x+40=0`

`<=>x^2-5x-8x+40=0`

`<=> x (x-5) - 8(x-5)=0`

`<=>(x-5)(x-8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=8\end{matrix}\right.\)

Answer 2

[Sáng]

5. \(x^2-x=12\Leftrightarrow x^2-x-12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

6. \(2x^2-3x=15-4x\Leftrightarrow2x^2+x-15=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

7. \(x\left(x-5\right)=24\Leftrightarrow x^2-5x-24=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

8. \(x\left(x-3\right)=10\left(x-4\right)\Leftrightarrow x^2-3x=10x-40\)

\(\Leftrightarrow x^2-13x+40=0\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)