Question

giúp mình với ạ : mình đang cần gấp

1. rút gọn

A) x(x+1)nhân(x-1)-(x² -1)(x+1)

b)(x+1)nhân(x-2)-(2x-1)nhân(x+2)+2x nhân(x-1)

c) (x²+2x-1)nhân (x+2)-(x-1)nhân (2x+1)

2. tìm x

a)(x-1)nhân (x-3)-(x+3)nhân(x-1)=0

b ) (2x+1 )nhaan(x-1)-(x+1)nhân (2x-3)=7x-1

c) (x=2)nhân(x-3)-(x-3)nhân(x+5 )=-x+3

d) 2(x+1)nhân (x-3)-(2x+1) nhân(x-2 )=-2x-3

cảm ơn nhiều ạ

Answer 1

Bài 1 :

a) \(x\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x+1\right)\)

\(=\left(x^3-x\right)-\left(x^3+x^2-x-1\right)\)

\(=x^3-x-x^3-x^2+x+1\)

\(=1-x^2\)

b) \(\left(x+1\right)\left(x-2\right)-\left(2x-1\right)\left(x+2\right)+2x\left(x-1\right)\)

\(=\left(x^2-x+2\right)-\left(2x^2+3x-2\right)+\left(2x^2-2x\right)\)

\(=x^2-x+2-2x^3-3x+2+2x^3+2x\)

\(=x^2-2x+4\)

\(=\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\)

c) \(\left(x^2+2x-1\right)\left(x+2\right)-\left(x-1\right)\left(2x+1\right)\)

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

\(=x^3+4x^2+3x-2-2x^3+x+1\)

\(=-x^3+4x^2+4x-1\)

Answer 2

Bài 1

\(a)x\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x+1\right)\\ =\left(x+1\right)\left[x\left(x-1\right)-\left(x^2-1\right)\right]\\ =\left(1+x\right)\left(x^2-x-x^2+1\right)\\ =\left(1+x\right)\left(1-x\right)\\ =1-x^2\)

\(b)\left(x+1\right)\left(x-2\right)-\left(2x-1\right)\left(x+2\right)+2x\left(x-1\right)\\ =x^2-2x+x-2-\left(2x^2+4x-x-2\right)+2x^2-2x\\ =x^2-2x+x-2-(2x^2+3x-2)+2x^2-2x\\ =x^2-2x+x-2-2x^2-3x+2+2x^2-2x\\ =x^2-6x\)

\(c)\left(x^2+2x-1\right)\left(x+2\right)-\left(x-1\right)\left(2x+1\right)\\ =x^3+2x^2+2x^2+4x-x-2-\left(2x^2+x-2x-1\right)\\ =x^3+2x^2+2x^2+4x-x-2-\left(2x^2-x-1\right)\\ =x^3+2x^2+2x^2+4x-x-2-2x^2+x+1\\ =x^3+2x^2+4x-1\)

Answer 3

Bài 2

Answer 4

Bài 2 :

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

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

\(\Leftrightarrow\left(x-1\right)2x=0\)

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

Answer 5

2)

c) \(\left(x+2\right)\left(x-3\right)-\left(x-3\right)\left(x+5\right)=-x+3\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)-\left(x-3\right)\left(x+5\right)-\left(3-x\right)=0\)

\(\Leftrightarrow x^2-x-6-\left(x^2+2x-15\right)-\left(3-x\right)=0\)

\(\Leftrightarrow x^2-x-6-x^2-2x+15-3+x=0\)

\(\Leftrightarrow-2x+6=0\)

\(\Leftrightarrow-2x=-6\Rightarrow x=3\)