Phan cả PHát - stupid lv max 

Try a=3;b=8 or a=4;b=5 or a=5;b=4 

à mà thôi ko cần giải đâu mik nghĩ ra rồi

Vì \(1+\frac{1}{a}\ge2\forall a>0\) (1)

    \(1+\frac{1}{b}\ge2\forall b>0\)(2)

Từ (1) và (2) 

\(\Rightarrow\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge2.2=4\)(Trái với gt đề bài) 

Suy ra không có cặp số nguyên dương a,b nào thỏa mãn 

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=\frac{3}{2}\Leftrightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=\frac{3}{2}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=\frac{1}{2}\)

\(\Leftrightarrow\frac{a+b+1}{ab}=\frac{1}{2}\Leftrightarrow2\left(a+b+1\right)=ab\Leftrightarrow2a+2b+2-ab=0\)

\(\Leftrightarrow2a-ab-4+2b+6=0\Leftrightarrow a\left(2-b\right)-2\left(2-b\right)=-6\)

\(\Leftrightarrow\left(a-2\right)\left(2-b\right)=-6\)

Đến đây chắc dễ rồi

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Các bạn ơi giải giúp mink vs mink đg cần gấp

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Ta có \(a+b+c\ge3\sqrt[3]{abc}=3\)

Áp dụng bđt cosi ta có:

\(\frac{a^3}{\left(b+1\right)\left(c+2\right)}+\frac{b+1}{12}+\frac{c+2}{18}\ge3\sqrt[3]{\frac{a^3}{12.18}}=\frac{a}{2}\)

Làm tương tự

=>\(VT+\left(\frac{a+1}{12}+\frac{a+2}{18}\right)+\left(\frac{b+1}{12}+\frac{b+2}{18}\right)+\left(\frac{c+1}{12}+\frac{c+2}{18}\right)\ge\frac{a+b+c}{2}\)

=> \(VT\ge\frac{13}{36}.\left(a+b+c\right)-\frac{7}{12}\ge\frac{13}{36}.3-\frac{7}{12}=\frac{1}{2}\)(ĐPCM)

ko chi tiết lắm

(d) qua A(5; 6) : y = mx - 5m + 6 (1) 
(C) : (x - 1)² + (y - 2)² = 1 (2) 
Thay y từ (1) vào (2) ta có phương trình hoành độ giao điểm của (d) và (C) 
(x - 1)² + (mx - 5m + 4)² = 1 
Khai triển ra pt bậc 2 : (m² + 1)x² - 2(5m² - 4m + 1)x + 25m² - 40m + 17 = 0 (*) 
Để (d) tiếp xúc (C) thì (*) phải có nghiệm kép 
∆' = (5m² - 4m + 1)² - (m² + 1)(25m² - 40m + 17) = - 4(3m² - 8m + 4) = 4(m - 2)(2 - 3m) = 0 => m = 3/2; m = 2 
KL : Có 2 đường thẳng cần tìm 
(d1) : y = (3/2)(x - 1) 
(d2) : y = 2x - 4 

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Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2}+1\ge2\sqrt{\frac{1}{a^2}\cdot1}=\frac{2}{a}\)

\(\frac{1}{b^2}+2\ge2\sqrt{\frac{1}{b^2}\cdot2}=\frac{2\sqrt{2}}{b}\)

\(\frac{1}{c^2}+8\ge2\sqrt{\frac{1}{c^2}\cdot8}=\frac{2\sqrt{8}}{c}\)

Nhân theo vế 3 BĐT trên ta có:

\(VT\ge\frac{2}{a}\cdot\frac{2\sqrt{2}}{b}\cdot\frac{2\sqrt{8}}{c}=\frac{32}{abc}=VP\)

Khi \(a=1;b=\frac{1}{\sqrt{2}};c=\frac{1}{2\sqrt{2}}\)

a)  Điều kiện :  \(a\ne-b;b\ne1;a\ne-1\)

\(P=\frac{a^2\left(1+a\right)-b^2\left(1-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^3+a^2+b^3-b^2-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a+b\right)\left(a-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^2+b^2-a^2b^2+a-b-ab}{\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^2\left(1-b^2\right)-\left(1-b^2\right)+a\left(1-b\right)+\left(1-b\right)}{\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{\left(1-b\right)\left(a^2+a^2b-1-b+a+1\right)}{\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^2+a^2b+a-b}{1+a}\)

\(P=\frac{a\left(a+1\right)+b\left(a-1\right)\left(a+1\right)}{1+a}\)

\(P=\frac{\left(a+1\right)\left(a+ab-b\right)}{1+a}\)

P = a + ab - b

b)

P = 3

<=>  a + ab - b = 3

<=>  a(b+1) - (b+1) +1 - 3 = 0

<=>   (b+1)(a-1)  = 2

Ta có bảng sau với a, b nguyên

Vậy (a;b) \(\in\){ (3; 0) ; (0; -3)}

\(3\left(2a^2+b^2\right)=\left(1^2+1^2+1^2\right)\left(a^2+a^2+b^2\right)\ge\left(a+a+b\right)^2=\left(2a+b\right)^2\)

\(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)

\(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)

\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)

\(gt\rightarrow7\left(x^2+y^2+z^2\right)=6\left(xy+yz+zx\right)+2015\)

\(\Leftrightarrow7\left(x+y+z\right)^2=20\left(xy+yz+zx\right)+2015\)

Ta có: \(3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2\)

\(\Rightarrow7\left(x+y+z\right)^2\le\frac{20}{3}\left(x+y+z\right)^2+2015\)

\(\Leftrightarrow\frac{1}{3}\left(x+y+z\right)^2\le2015\)

\(\Leftrightarrow x+y+z\le\sqrt{6045}\)

\(P\le\frac{1}{3}\left(x+y+z\right)\le\frac{\sqrt{6045}}{3}\)

Dấu bằng xảy ra khi \(x=y=z=\frac{\sqrt{6045}}{3}\)hay \(a=b=c=\left(\frac{\sqrt{6045}}{3}\right)^{-1}\)