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

\(=x^3-x^2+x^2-x+x-1-x^3+2x^2-2x^2+4x-4x+8+2x^2-8x+8\)

\(=2x^2-8x+15\)

Thay x= 102 vào E, ta được:

\(2\cdot102^2-8\cdot102+15=2\cdot102\left(102-4\right)+15=204\cdot98+15=19992+15=20007\)

hằng đẳng thức có phải nhanh hơn ko -_-
 

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

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

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

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

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

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

\(p=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(p=\frac{1}{2}\left(5^{32}-1\right)=\frac{5^{32}-1}{2}\)

1) a thỏa mãn: a² + a + 1 = 0, rõ ràng a khác 0. Chia cả 2 vế cho a ta được: \(a+\frac{1}{a}=-1\)

• Mặt khác ta có: \(\left(a+\frac{1}{a}\right)^3=-1\Rightarrow a^3+3\cdot\left(a+\frac{1}{a}\right)+\frac{1}{a^3}=-1\Rightarrow a^3+\frac{1}{a^3}=2\)
• \(\Rightarrow\left(a^3+\frac{1}{a^3}\right)^2=4\Rightarrow a^6+\frac{1}{a^6}=2\)\(\Rightarrow\left(a^6+\frac{1}{a^6}\right)\left(a^3+\frac{1}{a^3}\right)=4\Rightarrow a^9+\frac{1}{a^9}+a^3+\frac{1}{a^3}=4\Rightarrow a^9+\frac{1}{a^9}=2\)
• ... \(\Rightarrow a^{3k}+\frac{1}{a^{3k}}=2\)
• \(\Rightarrow a^{2013}+\frac{1}{a^{2013}}=2\)

2) Từ: \(x^2+x^2y^2-2y=0\Rightarrow x^2\left(y^2+1\right)=2y\Rightarrow x^2=\frac{2y}{y^2+1}\)

Với mọi y thì: \(\left(y-1\right)^2\ge0\Leftrightarrow2y\le y^2+1\Leftrightarrow\frac{2y}{y^2+1}\le1\)Do đó \(x^2=\frac{2y}{y^2+1}\le1\Rightarrow-1\le x\le1\)(1)

Mặt khác: \(x^3+2y^2-4y+3=0\Leftrightarrow x^3+1+2\left(y-1\right)^2=0\)(2)

Từ (1) => \(x^3+1\ge0\forall x\Rightarrow VT\left(2\right)\ge VP\left(2\right)\forall x;y\)

Để TM (2) thì dấu "=" xảy ra, khi đó x = -1; y = 1

và suy ra \(Q=x^2+y^2=2\)

= (3x + 1 - x - 1)(3x + 1 + x + 1)

= 2x(4x + 2)

Em áp dụng hđt số 3 trong sgk nhé.

(3x+1+x+1) (3x+1-x-1) = (4x+2) 2x

                                      = 4x(2x+1)

bằng 0 đúng không mấy bạn???

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)

vân toc cano là v ta có; 

4(v + 2) = 5(v-2)

v = 18km/h

S = 4(v+2) = 80km

gọi quãng đường AB là x

ta kẻ bảng sau :

vì vận tốc dòng nước chảy là 2 km/h ta có pt:

x/4 - 2 = x/5 + 2

<=> 5x - 40 = 4x + 40

<=> 5x - 4x = 40 + 40

<=> x = 80

vậy quãng đường AB dài 80 km