Nhẩm nghiệm, thấy x=-1 thỉ P=0, phân tích đa thức dần thành nhân tử

P(x)=\(\left(x+1\right)\left(2x^3-9x^2+7x+6\right)\)

=\(2x^{^{ }4}+2x^3-9x^3-9x^2+7x^2+7x+6x+6\)

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

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

Đây là 1 tích trong đó có 3 số nguyên lien tiep.

Trong 3 so nguyen lien tiep co it nhat 1 so chan va 1 so chia het cho 3

=> h cua chung chia het cho 2x3=6.

Vay P chia het cho 6.

                                                                                                                                                                                                    bạn ơi h là j thế 

Xét \(\left(7x+1\right)^2-\left(x+7\right)^2-48\left(x^2-1\right)\)

\(=49x^2+14x+1-x^2-14x-49-48x^2+48\)

\(=0\)

Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)

a. Vì

1/2<2/3

3/4<4/5

.........

99/100<100/101 nên M<N

b.M.N=\(\frac{1.2.3.4......100}{2.3.4.5......101}\)=\(\frac{1}{101}\)

Ta có : 1/x - 1/(x+1) = 1/x(x+1)

<=> pcm \(\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

<=> pcm \(\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

<=> pcm 1/x(x+1) = 1/x(x+1)

Đây là điều luôn đúng nên ta có điều phải chứng minh

Chú ý : Chữ pcm là phải chứng minh

Ta có : \(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x+5}\)

\(=\frac{1}{x\left(x+1\right)}+\frac{1}{x^2+x+2x+2}+\frac{1}{x^2+2x+3x+6}+\frac{1}{x^2+3x+4x+12}+\frac{1}{x^2+4x+5x+20}+\frac{1}{x+5}\)

\(=\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)+2\left(x+1\right)}+\frac{1}{x\left(x+2\right)+3\left(x+2\right)}+\frac{1}{x\left(x+3\right)+4\left(x+3\right)}\)

\(+\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x+5}\)

\(=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)

Áp dụng chứng minh trên ta có :

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)

=1/x

+)\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

+)\(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x+5}\)

\(=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)

\(=\frac{1}{x}\)

a) \(P\left(x\right)=2x^2-7x^3-2x^2+13x+6\)

\(\Leftrightarrow P\left(x\right)=-7x^3+13x+6\)

\(\Leftrightarrow P\left(x\right)=-7x^3+7x+6x+6\)

\(\Leftrightarrow P\left(x\right)=-\left(7x^3-7x\right)+\left(6x+6\right)\)

\(\Leftrightarrow P\left(x\right)=-7x\left(x^2-1\right)+6\left(x+1\right)\)

\(\Leftrightarrow P\left(x\right)=-7x\left(x-1\right)\left(x+1\right)+6\left(x+1\right)\)

\(\Leftrightarrow P\left(x\right)=\left(x+1\right)\left[-7x\left(x-1\right)+6\right]\)

\(\Leftrightarrow P\left(x\right)=\left(x+1\right)\left(-7x^2+7x+6\right)\)

mink làm câu b nha

Từ câu a ta có

P(x)=-7x(x-1)(x+1)+6(x+1)

=>P(x)=6(x+1)-7x(x-1)(x+1)

Vì x(x-1)(x+1) chia hết cho 6 ( Đã đc chứng minh)

=>7x(x-1)(x+) chia hết cho 6

mà 6(x+1) chia hết cho 6

=> 6(x+1) -7x(x-1)(x+1) chia hết cho 6

=> P(x) chia hết cho 6