Hướng dẫn

Đặt là x,y,z

Chứng minh được là \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

- 1 hay + 1 v?

+ 1 hợp lí hơn

\(A=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt : \(a^2+8+11=t\) khi đó pt trở thành :

\(\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)

\(=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)

\(=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

Chúc bạn học tốt !!!

A = (a+1)(a+3)(a+5)(a+7) + 15

A = [ (a+1) (a+7)] [(a+3) (a+5)] + 15

A= ( a² + 8a + 7)( a² + 8a + 15 ) + 15                 (*)

         Đặt a² + 8a + 7 = t

=> A = t.(t+8) + 15

     A = t² + 8t + 15

     A = t² + 3t + 5t + 15

     A = ( t +3).(t+5)

  Thay   A = ( t +3).(t+5) vào (*)

=> A = ( a² + 8a + 7 + 3).( a² + 8a + 7 + 5)

    A = ( a² + 8a + 10).( a² + 8a + 12 )

     A = ( a² + 8a + 10).( a² + 6a + 2a + 12 )

       A = ( a² + 8a + 10) ( a+6)(a+2)

\(-3x^2+4x-2020\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{2020}{3}\right)\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}+\frac{6056}{9}\right)\)

\(=-3\left[\left(x-\frac{2}{3}\right)^2+\frac{6056}{9}\right]\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{6056}{3}\ge-\frac{6056}{3}\)

(Dấu "=" \(\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\))

A B C D E

Sao BC cắt DE được nhỉ ?

phân tích đa thức trên thành nhân tử

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

\(c)4x\left(x-2\right)-\left(2-x\right)^2=4x\left(x-2\right)+\left(x-2\right)^2=\left(4x+x-2\right)\left(x-2\right)=\left(5x-2\right)\left(x-2\right)\)

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

c) \(4x\left(x-2\right)-\left(2-x\right)^2\)

\(=4x\left(x-2\right)-\left(x-2\right)^2\)

\(=\left(x-2\right)\left(4x-x+2\right)\)

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

d) \(\left(x-2\right)^2-\left(2-x\right)^3\)

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

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

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

Từ giả thiết => $a\equiv 1\left(mod3\right)$, a=3k+1 ($k\in N$); b$\equiv 2\left(mod3\right)$, b=3q+2 $\left(q\in N\right)$

=> $A=4^a+9^b+a+b=1=1+0+1+2\left(mod3\right)$hay $A\equiv 4\left(mod3\right)$(1)

Lại có $4^a=4^{3k+1}=4\cdot {64}^k\equiv 4\left(mod7\right)$

$9^b=9^{3q+2}\equiv 2^{3q+2}\left(mod7\right)\Rightarrow 9^b\equiv 4\cdot 8^q\equiv 4\left(mod7\right)$

Từ gt => $a\equiv 1\left(mod7\right),b\equiv 1\left(mod7\right)$

Dẫn đến $A=4^a+9^b+a+b\equiv 4+4+1+1\left(mod7\right)$hay $A\equiv 10\left(mod7\right)$

Từ (1) => $A\equiv 10\left(mod3\right)$mà 3,7 nguyên tố cùng nhau nên $A\equiv 10\left(mod21\right)$

=> A chia 21 dư 10

Từ giả thiết => $a\equiv 1\left(mod3\right)$, a=3k+1 ($k\in N$); b$\equiv 2\left(mod3\right)$, b=3q+2 $\left(q\in N\right)$

=> $A=4^a+9^b+a+b=1=1+0+1+2\left(mod3\right)$hay $A\equiv 4\left(mod3\right)$(1)

Lại có $4^a=4^{3k+1}=4\cdot {64}^k\equiv 4\left(mod7\right)$

$9^b=9^{3q+2}\equiv 2^{3q+2}\left(mod7\right)\Rightarrow 9^b\equiv 4\cdot 8^q\equiv 4\left(mod7\right)$

Từ gt => $a\equiv 1\left(mod7\right),b\equiv 1\left(mod7\right)$

Dẫn đến $A=4^a+9^b+a+b\equiv 4+4+1+1\left(mod7\right)$hay $A\equiv 10\left(mod7\right)$

Từ (1) => $A\equiv 10\left(mod3\right)$mà 3,7 nguyên tố cùng nhau nên $A\equiv 10\left(mod21\right)$

=> A chia 21 dư 10

\(4a^2-x^2-2x-1\)

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

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

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

Ta có : 4a² - x² - 2x - 1

        = 4a² - ( x² + 2x + 1 )

       = ( 2a )² - ( x + 1 )²

       =  ( 2a - x - 1 )( 2a + x + 1 )