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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 8: a) -1/2x^2 (2x^3-4x+3) b)3x.(x-2)+2.(x-2)=0 c)x.(2x-1)+4x=2 giúp mình với ạ:(

Toán Lớp 8: a) -1/2x^2 (2x^3-4x+3)
b)3x.(x-2)+2.(x-2)=0
c)x.(2x-1)+4x=2
giúp mình với ạ:(

1. Giải đáp:

Lời giải và giải thích chi tiết:
a) -1/[2x^2] (2x^3 -4x+3)
=-x+2/x -3/[2x^2]
b)3x.(x-2)+2.(x-2)=0
=>(x-2)(3x+2)=0
=>$$\left[ \begin{array}{l}x-2=0\\3x+2=0\end{array} \right.$$ $\\$=>$$\left[ \begin{array}{l}x=2\\x=-\dfrac{2}{3}\end{array} \right.$$
c)x.(2x-1)+4x=2
=>x(2x-1)+4x-2=0
=>x(2x-1)+2(2x-1)=0
=>(2x-1)(x+2)=0
=>$$\left[ \begin{array}{l}2x-1=0\\x+2=0\end{array} \right.$$ $\\$=>$$\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-2\end{array} \right.$$
(Bài tham khảo)

2. a) \frac{-1}{2}x^2 ( 2x^3 – 4x + 3 )
= -x^5 + 2x^3 – \frac{3}{2}
b) 3x ( x – 2 ) + 2 ( x – 2 ) =0
⇒ ( x – 2 ) ( 3x + 2 ) =0
⇒ $$\left[ \begin{array}{l}x-2=0\\3x+2=0\end{array} \right.$$
⇒ $$\left[ \begin{array}{l}x=2\\x=-\dfrac{2}{3}\end{array} \right.$$
Vậy x ∈ { 2 ; \frac{-2}{3} }
c) x ( 2x – 1 ) + 4x = 2
⇒ x ( 2x – 1 ) + 4x – 2 =0
⇒ x ( 2x – 1 ) + 2 ( 2x – 1 )=0
⇒ ( x + 2 ) ( 2x – 1 )=0
⇒ $$\left[ \begin{array}{l}x+2=0\\2x-1=0\end{array} \right.$$
⇒ $$\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{2}\end{array} \right.$$
Vậy x ∈ { -2 ; \frac{1}{2} }