a) Ta có: 

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

=> Đpcm 

b) 

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

đáp án : x-5y dư -10y^2+5

`a) 2024x - 2024y - 2024x^2 + 4048xy - 2024y^2`

`= ( 2024x - 2024y) - (2024x^2 - 4048xy + 2024y^2)`

`= 2024 (x-y) - 2024 (x^2 - 2xy + y^2)`

`= 2024 (x-y) - 2024 (x-y)^2`

`= 2024 (x-y) (1 - x + y)`

`b) x^2 + 5x - 6`

`= (x^2 - x) + (6x - 6)`

`= x(x-1) + 6(x-1)`

`= (x+6)(x-1)`

`c) 2x^2 + 3x - 5`

`= (2x^2 - 2x) + (5x - 5)`

`= 2x(x - 1) + 5(x-1)`

`= (2x+5)(x-1)`

`d) x^4 + 4 `

`= (x^2)^2 + 2^2`

`= (x^2)^2 + 4x^2 + 2^2 - 4x^2`

`= (x^2 + 2)^2 - (2x)^2 `

`= (x^2 - 2x + 2)(x^2 + 2x+ 2)`

`e) x^5 + x + 1`

`= (x^5 - x^2) + (x^2 + x + 1)`

`= x^2 (x^3 - 1) + (x^2 + x + 1)`

`= x^2 (x-1) (x^2 + x + 1) + (x^2 + x + 1)`

`= (x^3 - x^2 + 1) (x^2 + x + 1) `

cuu

Yêu cầu của đề bài là gì vậy bạn?

a: \(101^2=\left(100+1\right)^2=100^2+2\cdot100\cdot1+1^2\)

=10000+200+1

=10201

b: \(64^2+36^2+72\cdot64\)

\(=64^2+2\cdot64\cdot36+36^2\)

\(=\left(64+36\right)^2=100^2=10000\)

c: \(54^2+46^2-2\cdot54\cdot46=\left(54-46\right)^2=8^2=64\)

d: \(98\cdot102=\left(100-2\right)\left(100+2\right)=100^2-4=9996\)

\(3\cdot\left(2\cdot1\right)=6x-3x\\ 6=3x\\ x=\dfrac{6}{3}=2\)

`= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2024(x^2 + x + 1)`

`= (x^2 - x + 2024)(x^2 + x + 1)`.

\(x^4+6x^3+7x^2-6x+1\\ =\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)