Làm này mới đúng chứ:

(3x-5)(x+1) - (x-1)(x+2)= (2x-3)(x+2) + 1

=> (3x-5)(x+1) = (2x-3)(x+2) + (x-1)(x+2) + 1

=> (3x - 4 - 1)(x + 1) = (x+2) [(2x-3) + (x-1)] + 1

=> (3x-4)(x+1) - x - 1 = (x+2).(3x-4) + 1

=> (3x-4)(x+1) - (x+2).(3x-4) = x + 1 + 1

=> (3x-4).[(x+1) - (x+2)] = x + 2

=> (3x-4).(-1) = x + 2

=> - 3x + 4 = x + 2

=> 3x + x = 4 + 2

=> 4x = 6

=> x = 6 : 4

=> x = 3/2

thế bn nhan vào dj,rồi giải ra là đc

Phạm Tuấn Kiệt bn lm sai kết quả phải là 1/2 chứ

bn thử lấy kết quả của bn tính mà xem ko đúng đâu

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

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

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

b) Ta có: \(\left(x^3-x^2+2x-1\right)\left(5-x\right)\)

\(=5x^3-x^4-5x^2+x^3+10x-2x^2-5+5x\)

\(=-x^4+6x^3-7x^2+15x-5\)

Ta có: \(\left(x-5\right)\left(x^3-x^2+2x-1\right)\)

\(=-\left(5-x\right)\left(x^3-x^2+2x-1\right)\)

\(=x^4-6x^3+7x^2-15x+5\)

`(2x-3)(x^2-5+1)`

`=2x(x^2-5+1)-3(x^2-5+1)`

`= 2x*x^2+2x*(-5)+2x-[3*x^2+3*(-5)+3]`

`= 2x^3-10x+2x-(3x^2-15+3)`

`= 2x^3-10x+2x-3x^2+15-3`

`= 2x^3-3x^2-8x+12`

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

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

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

(x-1)(x\(^5\)+x\(^4\)+x\(^3\)+x\(^2\)+x+1)

=x(\(x^5+x^4+x^3+x^2+x+1\))-1(\(x^5+x^4+x^3+x^2+x+1\))

= x.\(x^5+x\cdot x^4+x\cdot x^3+x\cdot x^2+x\cdot x+x\cdot1\)-1.\(x^5-1\cdot x^4-1\cdot x^3-1\cdot x^2-1\cdot x-1\cdot1\)

=\(x^6\)+\(x^5\)\(+x^4\)+\(x^3\)+\(x^2\)+1x -1\(x^5\)-1\(x^4\)-1\(x^3\)-1\(x^2\)-1x -1

=\(x^6\)+(\(x^5\)-1\(x^5\))+(x\(^4\)-1\(x^4\))+(\(x^3\)-1\(x^3\))+(x\(^2\)-1\(x^2\))+(1x-1x)-1

=x\(^6\)-1

thanks

a) (x-1)*(x+2)-(x-3)*(-x+4)=19

\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)

\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)

\(\Leftrightarrow2x^2-3x+7-19=0\)

\(\Leftrightarrow2x^2-3x-12=0\)

Đề sai??

b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)

\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)

\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)

\(\Leftrightarrow-30x^2+7x-4+17=0\)

\(\Leftrightarrow-30x^2+7x+13=0\)

???

c) (x+1)*(x+1)-(x-1)*(x-1)=9

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=9\)

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

\(\Leftrightarrow2x.2=9\)

\(\Leftrightarrow x=\frac{9}{4}\)

\(Câu8\)

\(a,A=\dfrac{1}{2}x^3\times\dfrac{8}{5}x^2=\left(\dfrac{1}{2}\times\dfrac{8}{5}\right)x^{3+2}=\dfrac{4}{5}x^5\)

b, \(P\left(0\right)=0^2-5.0+6=6\\ P\left(2\right)=2^2-5.2+6=0\)

Câu 9

\(a,A\left(x\right)+B\left(x\right)=5x^3+x^2-3x+5+5x^3+x^2+2x-3\\ =\left(5x^3+5x^3\right)+\left(x^2+x^2\right)+\left(-3x+2x\right)+\left(5-3\right)\\ =10x^3+2x^2-x+2\)

\(b,H\left(x\right)=A\left(x\right)-B\left(x\right)=5x^3+x^2-3x+5-\left(5x^3+x^2+2x-3\right)\\ =5x^3+x^2-3x+5-5x^3-x^2-2x+3\\ =\left(5x^3-5x^3\right)+\left(x^2-x^2\right) +\left(-3x-2x\right)+\left(5+3\right)\\ =-5x+8\)

\(H\left(x\right)=0\\ \Rightarrow-5x+8=0\\ \Rightarrow x=\dfrac{8}{5}\)

vậy nghiệm của đa thức là \(x=\dfrac{8}{5}\)