đáp án : x-5y dư -10y^2+5

• 1,She prefers drinking a cup of coffee.
• 2,We would prefer going abroad on a long vacation.
• 3,Why do you prefer to attend a concert rather than watch it live?
• 4,I would prefer to work part-time rather than work full-time.
• 5,I like both Harry Potter and Peaky Blinders. But I prefer Harry Potter.

1 prefers

2 prefers

3 prefer

5 would prefer

5 prefer

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) 

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

\(\dfrac{3001}{5002}=1-\dfrac{2001}{5002}< 1\)

\(\dfrac{444111}{666111}=\dfrac{4001\times111}{6001\times111}=\dfrac{4001}{6001}=1-\dfrac{2000}{6001}< 1\)

\(\dfrac{30303040404}{505050}>1\)

Ta có: \(\dfrac{2001}{5002}>\dfrac{2000}{5002}>\dfrac{2000}{6001}\)

=> \(1-\dfrac{2001}{5002}< 1-\dfrac{2000}{6001}\)

=> \(\dfrac{3001}{5002}< \dfrac{444111}{666111}\)

Vậy:

Thứ tự sắp xếp từ bé đến lớn là: 

\(\dfrac{3001}{5002};\dfrac{444111}{666111};\dfrac{30303040404}{505050}\)

Thứ tự sắp xếp từ lớn đến bé là: 

\(\dfrac{30303040404}{505050};\dfrac{444111}{666111};\dfrac{3001}{5002}\)

\(B=4^{31}-1\\ =\left(4^{16}-1\right)\left(4^{16}+1\right)\\ =\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\\ =\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\\ =\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\\ =\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\\ =3\left[\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\right]\\ =3A\)

1 A. Not to go
2 B. Than
3 A. Prefer
4 B. Picked

1 C

2 B

3 A

4 B

shop ơi shop giải được chưa ạ cứu em bài đó

Đề của bạn cho là phương trình, không phải đa thức. Bạn xem lại nhé.

Dạ

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

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

=>\(a^2b+a^2c-b^2a-b^2c=0\)

=>\(ab\left(a-b\right)+c\left(a^2-b^2\right)=0\)

=>ab(a-b)+c(a-b)(a+b)=0

=>(a-b)(ab+ac+bc)=0

=>ab+ac+bc=0(do a-b<>0)

=>c(a+b)=-ab

=>\(a+b=-\frac{ab}{c}\)

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