Question

Tìm x

Y) x^2-x-6=0

Z) 3x² –5x–8=0

J) 25x^2-4=0

 R) 2(x+3)-x^2-3x=0

 U. x³–3x² –x+3=0 

Giúp mik vs mình cần gấp 

Answer 1

Toán mik ghi nhầm ko phải ta

Answer 2

em ghi nhầm môn em có thể đăng lại ko em

Answer 3

y) \(x^2-x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{-2;3\right\}\) là nghiệm của pt.

z) \(3x^2-5x-8=0\\ \Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{8}{3};-1\right\}\) là nghiệm của pt.

j) \(25x^2-4=0\\ \Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{2}{5};\dfrac{-2}{5}\right\}\) là nghiệm của pt.

r) \(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{-3;2\right\}\) là nghiệm của pt.

u) \(x^3-3x^2-x+3=0\\ \Leftrightarrow x^2\left(x-3\right)-\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{-1;1;3\right\}\) là nghiệm của pt.

 

Answer 4

y)x²-x-6=0

⇔(x-3)(x+2)=0

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

z)3x²-5x-8=0

⇔ (3x-8)(x+1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

j)25x²-4=0

⇔(5x-4)(5x+4)=0

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

r)2(x+3)-x²-3x=0

⇔2(x+3)-x(x+3)=0

⇔ (x+3)(2-x)=0

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

u)x³-3x²-x+3=0

⇔x²(x-3)-(x-3)=0

⇔(x-3)(x²-1)=0

⇔(x-3)(x-1)(x+1)=0

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

Answer 5

y: Ta có: \(x^2-x-6=0\)

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

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

z: Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

j: Ta có: \(25x^2-4=0\)

\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)

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