1 với 0 nhé

Ta có:
p = 2 ➙ \(2^{p}\) + p² = 8 (hợp số) (Loại)
p = 3 ➙ \(2^{p}\) + p² = 17 (số nguyên tố) (Nhận)
p > 3 ➙ \(2^{p}\) + p² = (\(2^{p}\) + 1) + (p² - 1)
Vì p lẻ và p không chia hết cho 3, nên:
\(2^{p}\) + 1 ⋮ 3 và p² - 1 ⋮ 3
➜ \(2^{p}\) + p² ⋮ 3 (hợp số) (Loại)

Vậy với p = 3 thì \(2^{p}\) + p² cũng là số nguyên tố.

từ ioe à

1 A

2 D

3 B

4 D

5 B

Olm chào em, em xem hướng dẫn chi tiết dưới đây em sẽ hiểu vì sao em nhé.

Giải:

\(x^2\) - 5\(x\) + 6

= (\(x^2\) - 3\(x\)) - (2\(x-6\))

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

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

Vì (x-3) (x-2)=x^2-5x+6

1: Để (d) cắt (d') tại một điểm trên trục tung thì

\(\left\{{}\begin{matrix}a\ne a'\\b=b'\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m\ne m-2\\m-1=-2m\end{matrix}\right.\Leftrightarrow3m=1\)

=>\(m=\dfrac{1}{3}\)

2: Thay x=2 vào y=mx+m-1, ta được:

\(y=m\cdot2+m-1=3m-1\)

Thay x=2 và y=3m-1 vào (d'), ta được:

\(2\left(m-2\right)-2m=3m-1\)

=>3m-1=-4

=>3m=-3

=>m=-1

3: Thay x=-1 và y=2 vào (d), ta được:

\(m\cdot\left(-1\right)+m-1=2\)

=>-m+m-1=2

=>-1=2(vô lý)

vậy: \(m\in\varnothing\)

a: 3x+22-3x+16=53+2x

=>2x+53=38

=>2x=38-53=-15

=>\(x=-\dfrac{15}{2}\)

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

=>\(\left(x+1\right)\left(x+2\right)-\left(2-x\right)\left(x+2\right)=0\)

=>\(\left(x+2\right)\left(x+1-2+x\right)=0\)

=>(x+2)(2x-1)=0

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

c: \(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a^2-2ab+b^2-4c^2\right)\)

\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)

\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)

d: \(x^2-25+y^2+2xy\)

\(=\left(x^2+2xy+y^2\right)-25\)

\(=\left(x+y\right)^2-25\)

=(x+y+5)(x+y-5)

Phân tích các đa thức sau thành nhân tử:

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

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

\(=\dfrac{x^2+x-1}{x^2+x+1}+\dfrac{-x^2+x+1}{x^2+x+1}+\dfrac{x^2-x+1}{x^2+x+1}\)

\(=\dfrac{x^2+x-1-x^2+x+1+x^2-x+1}{x^2+x+1}\)

\(=\dfrac{x^2+x+1}{x^2+x+1}=1\)

\(B=\dfrac{1}{a-b}+\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)

\(=\dfrac{a+b+a-b}{\left(a-b\right)\left(a+b\right)}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)

\(=\dfrac{2a}{a^2-b^2}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)

\(=\dfrac{2a\left(a^2+b^2\right)+2a\left(a^2-b^2\right)}{\left(a^2-b^2\right)\left(a^2+b^2\right)}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)

\(=\dfrac{4a^3}{a^4-b^4}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)

\(=\dfrac{4a^3\left(a^4+b^4+a^4-b^4\right)}{\left(a^4-b^4\right)\left(a^4+b^4\right)}+\dfrac{8a^7}{a^8+b^8}\)

\(=\dfrac{8a^7}{a^8-b^8}+\dfrac{8a^7}{a^8+b^8}\)

\(=\dfrac{8a^7\left(a^8-b^8+a^8+b^8\right)}{a^{16}-b^{16}}=\dfrac{16a^{15}}{a^{16}-b^{16}}\)

Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

=>\(a^2+b^2+c^2+2\left(ab+ac+bc\right)=a^2+b^2+c^2\)

=>2(ab+ac+bc)=0

=>2ab+2ac+2bc=0

=>ab+ac+bc=0

=>ab=-ac-bc; ac=-ab-bc; bc=-ab-ac

\(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac\)

\(=a^2-ab-ac+bc=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)

\(b^2+2ac=b^2+ac+ac=b^2+ac-ba-bc\)

\(=b^2-ba-bc+ac=b\left(b-a\right)-c\left(b-a\right)\)

=(b-c)(b-a)=-(a-b)(b-c)

\(c^2+2ab=c^2+ab+ab=c^2+ab-ac-bc\)

\(=c^2-ac-bc+ab=c\left(c-a\right)-b\left(c-a\right)=\left(c-a\right)\left(c-b\right)=\left(a-c\right)\left(b-c_{}\right)\)

\(P=\frac{2}{a^2+2bc}+\frac{2}{b^2+2ac}+\frac{2}{c^2+2ab}\)

\(=\frac{2}{\left(a-b\right)\left(a-c\right)}-\frac{2}{\left(a-b\right)\left(b-c\right)}+\frac{2}{\left(a-c\right)\cdot\left(b-c\right)}\)

\(=\frac{2\left(b-c\right)-2\left(a-c\right)+2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{2\left(b-c-a+c+a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)