Question

[Huy Bùi Quang]

1,Cho a + b = 1. TÍnh giá trị biểu thức A: a³ + b³ + 3a.b.(a² + b²)+6a² .b² .(a + b)

2,Rút gọn biểu thức sau:

(a + b + c)³- (b + c - a)³-(a + c - b)³ - (a + b - c)³

3,

Chứng minh rằng nếu p và p² + 8 là các sô nguyên tố thì p² +2 cũng là số nguyên tố

Cho a,b,c khác 0, thỏa mãn a+b+c=a.b.c và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\) .Chứng minh rằng \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)

Answer 1

[Nguyễn Huy Tú]

1, \(a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=a^3+b^3+3a^3b+3ab^3+6a^2b^2\)

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

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

\(=a^2-ab+b^2+3ab\)

\(=a^2+2ab+b^2=\left(a+b\right)^2\)

\(=1\)

Vậy A = 1

Bài 2: ( đặt đề bài là A )

Đặt \(b+c-a=x,a+c-b=y,a+b-c=z\)

\(\Rightarrow a+b+c=x+y+z\)

\(\Leftrightarrow A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)-x^3-y^3-z^3\)

\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(=3.2c.2a.2b=24abc\)

Vậy...

Bài 3:

+) Xét p = 3 có: \(p^2+2=11\in P\) ( t/m )

+) Xét \(p\ne3\) thì:

+ \(p=3k+1\Rightarrow p^2+2=\left(3k+1\right)^2+2=9k^2+6k+3⋮3\notin P\)

+ \(p=3k+2\Rightarrow p^2+2=\left(3k+2\right)^2+2=9k^2+12k+6⋮3\notin P\)

Vậy p = 3

Bài 4:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2c}{abc}+\dfrac{2a}{abc}+\dfrac{2b}{abc}=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)

\(\Rightarrowđpcm\)