Question

_{^{4(x^2 + x) - 19 =-10 - x}}

Answer 1

Answer 2

4(x² + x) - 19 = -10 - x

\(\Leftrightarrow\) 4x² + 4x - 19 + 10 + x = 0

\(\Leftrightarrow\) 4x² + 5x - 9 = 0

\(\Leftrightarrow\) 4x² - 4x + 9x - 9 = 0

\(\Leftrightarrow\) 4x(x - 1) + 9(x - 1) = 0

\(\Leftrightarrow\) (x - 1)(4x + 9) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-1=0\\4x+9=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\x=\dfrac{-9}{4}\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt!

Answer 3

Ta có: \(4\left(x^2+x\right)-19=-10-x\)

\(\Leftrightarrow4x^2+4x-19+10+x=0\)

\(\Leftrightarrow4x^2+5x-9=0\)

\(\Leftrightarrow4x^2+9x-4x-9=0\)

\(\Leftrightarrow x\left(4x+9\right)-\left(4x+9\right)=0\)

\(\Leftrightarrow\left(4x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};1\right\}\)

Answer 4

4.\(x^2\)+ 4.x - 19 = -10 - x

\(4.x^2\)+ 5.x - 9 = 0

Nên: \(\Delta=5^2-4.4.\left(-9\right)=169\)

Do đó: Phương trình có hai nghiệm phân biệt

x1 = \(\dfrac{-5+\sqrt{169}}{2.4}=1\)

x2 = \(\dfrac{-5-\sqrt{169}}{2.4}=-\dfrac{9}{4}\)

Vậy....

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