a.

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

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

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

b.

Từ câu a:

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

\(\Rightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

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

 A = (14⁸)²⁰²⁰ + 10 

A =  (14⁸)^5.404 + 10

A = (14⁵)^8.404 + 10

A = 537824³²³² + 10

537824 \(\equiv\) 1 (mod 11)

537824³²³² \(\equiv\) 1³²³² (mod 11) \(\equiv\) 1 (mod 11)

10 \(\equiv\) 10 (mod 11)

⇒ 537824³²³² + 10  \(\equiv\) 1 + 10 (mod 11)

⇒537824³²³² + 10 \(\equiv\) 11 (mod 11) \(\equiv\) 0 (mod 11)

⇒ A = (14⁸)²⁰²⁰ + 10 \(⋮\) 11 (đpcm)

\(14\equiv3\left(mod11\right)\Rightarrow\left(14^8\right)^{2020}\equiv\left(3^8\right)^{2020}\left(mod11\right)\)

\(\left(3^8\right)^{2020}=3^{8.404.5}=\left(3^5\right)^{3232}=\left(243\right)^{3232}\)

\(243\equiv1\left(mod11\right)\Rightarrow243^{3232}\equiv1\left(mod11\right)\)

\(\Rightarrow\left(14^8\right)^{2020}\equiv1\left(mod11\right)\)

\(\Rightarrow\left(14^8\right)^{2020}+10⋮11\)

g) (3x + 4)(3x - 4) - (2x + 5)² = (x - 5)² + (2x + 1)² - (x² - 2x) + (x - 1)²

9x² - 16 - 4x² - 20x - 25 = x² - 10x + 25 + 4x² + 4x + 1 - x² + 2x + x² - 2x + 1

5x² - 20x - 41 = 5x² - 6x + 27

5x² - 5x² - 20x + 6x = 27 + 41

-14x = 68

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

-5(x² + 6x + 9) + x² - 1 + 4x² - 12x + 9 = 25x² - 20x + 4 - 25x² - 15x

-5x² - 30x - 45 + x² - 1 + 4x² - 12x + 9 = -35x + 4

-42x - 37 = -35x + 4

-42x + 35x = 4 + 37

-7x = 41

a) (-x + 5)(x - 2) + (x - 7)(x + 7) = (3x + 1)² - (3x - 2)(3x + 2)

-x² + 2x + 5x - 10 + x² - 49 = 9x² + 6x + 1 - 9x² + 4

7x - 59 = 6x + 5

7x - 6x = 5 + 59

x = 64

b) (5x - 1)(x + 1) - 2(x - 3)² = (x + 2)(3x - 1) - (x + 4)² + (x² - x)

5x² + 5x - x - 1 - 2(x² - 6x + 9) = 3x² - x + 6x - 2 - x² - 8x - 16 + x² - x

5x² + 4x - 1 - 2x² + 12x - 18 = 3x² - 4x - 18

3x² + 16x - 19 = 3x² - 4x - 18

3x² + 16x - 3x² + 4x = -18 + 19

20x = 1

Phân tích nhân tử à bạn?

1

a.

\(\left(x+2y\right)^2-\left(x-2y\right)^2=\left(x+2y+x-2y\right)\left(x+2y-x+2y\right)=2x.4y=8xy\)

b.

\(\left(3x+2y\right)^2-\left(3x+2y\right)\left(6y-4x\right)+\left(2x-3y\right)^2\)

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

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

\(=\left(4x\right)^2=16x^2\)

\(A=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\)

Do \(\left(x-3\right)^2\ge0;\forall x\Rightarrow-\left(x-3\right)^2\le0;\forall x\)

\(\Rightarrow A\le0\Rightarrow A_{max}=0\) khi \(x=3\)

\(B=4x^2-4x+1+14=\left(2x-1\right)^2+14\)

Do \(\left(2x-1\right)^2\ge0;\forall x\Rightarrow\left(2x-1\right)^2+14\ge14;\forall x\)

\(\Rightarrow B_{min}=14\) khi \(2x-1=0\Rightarrow x=\dfrac{1}{2}\)

\(\left(x^2+4y^2-4xy\right):\left(x-2y\right)=\left(x-2y\right)^2:\left(x-2y\right)=x-2y\)

Chắc em ghi nhầm đề, hoặc là số giữa là \(2\left(2+3x\right)\left(1-2y\right)\), hoặc là số cuối là \(\left(2x-1\right)^2\)