Cho a - 4ab = b va a > b. Tinh gia tri cua bieu thuc P=a+b/a-b
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A=\(\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=\left(1-\frac{2}{5}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{2}{9}\right)\cdot...\cdot\left(1-\frac{2}{79}\right)=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{77}{79}=\frac{3}{79}\)
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cậu thế vào, Ta có:
x=12,y=5
Vậy x+y=17.Toán vòng 11 olympic chứ gì, mình thi rồi.
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\\ \Rightarrow yz+xz+xy=0\)
\(A=\frac{xy}{z^2}+\frac{xz}{y^2}+\frac{yz}{x^2}\\ \Leftrightarrow A=\frac{x^3y^3+x^3z^3+y^3z^3}{x^2y^2z^2}\)
Ta có :\(yz+xz+xy=0\)
\(\Rightarrow y^3x^3+x^3z^3+x^3y^3=-3xyz\left(y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz\right)\)
\(=-3xyz\left(yz+xz\right)\left(xz+xy\right)\left(yz+xy\right)\)
\(=-3xyz\left(-xy\right)\left(-yz\right)\left(-xz\right)\\ =3x^2y^2z^2\)
\(\Rightarrow A=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
Ta có a-4ab=b. suy ra: \(a=b+4ab\)
Suy ra: \(P=\frac{b+4ab+b}{b+4ab-b}=\frac{2b+4ab}{4ab}=1+\frac{1}{2a}\)