Ta có a-4ab=b. suy ra: \(a=b+4ab\)

Suy ra: \(P=\frac{b+4ab+b}{b+4ab-b}=\frac{2b+4ab}{4ab}=1+\frac{1}{2a}\)

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

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)

2.GTLN:4

1.x=-5

\(A=\left(1-\frac{2}{5}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{2}{9}\right)\cdot...\cdot\left(1-\frac{2}{79}\right)=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{77}{79}=\frac{3}{79}\)

3/79 đúng chắc, mình bấm máy cả buổi.

cậu thế vào, Ta có:

x=12,y=5

Vậy x+y=17.Toán vòng 11 olympic chứ gì, mình thi rồi.

5 và 12 bạn nhé 

Theo bài ra , ta có : 

x² - 6x + 8 =0 

(=) x² - 2x - 4x + 8 = 0 

(=) x(x-2) - 4(x-2) = 0 

(=) (x-2)(x-4) = 0 

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\)  \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

Vậy x=2;x=4

-6 nha bạn nhớ 

x(x+1)-x(x-3)=0

=>x²+x-x²+3x=0

=>4x=0

=>x=0

vậy x=0 thì x(x+1)-x(x-3)=0

\(xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\\ \Rightarrow yz+xz+xy=0\)

\(A=\frac{xy}{z^2}+\frac{xz}{y^2}+\frac{yz}{x^2}\\ \Leftrightarrow A=\frac{x^3y^3+x^3z^3+y^3z^3}{x^2y^2z^2}\)

Ta có :\(yz+xz+xy=0\)

         \(\Rightarrow y^3x^3+x^3z^3+x^3y^3=-3xyz\left(y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz\right)\)

                                                  \(=-3xyz\left(yz+xz\right)\left(xz+xy\right)\left(yz+xy\right)\)

                                                   \(=-3xyz\left(-xy\right)\left(-yz\right)\left(-xz\right)\\ =3x^2y^2z^2\)

      \(\Rightarrow A=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)