\(P=loga^3+logb^2=log\left(a^3b^2\right)=log\left(100\right)=10\)

Chọn A.

Ta có:

1 log a b - 1 log b a = 2018   ⇔ log a b + 1 log a b = 2018 ⇔ t + 1 t = 2018  

  P = 1 log ab b - 1 log ab a = log b ab - log a ab = log b a - log a b = 1 log a b - log a b = 1 t - t

Mà ( t + 1 t ) 2 - ( 1 t - t ) 2 = 4  suy ra

P = 1 t - t = ( t + 1 t ) 2 - 4 = 2018 - 4 = 2014 .

Đáp án B

Đáp án A

Từ \(a+b\ge1=>b\ge1-a>0\) ta có:

A = \(\dfrac{8a^2+b}{4a}+b^2\ge\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2\)

=\(\dfrac{8a^2-a+1+4a^3-8a^2+4a}{4a}=\dfrac{4a^3-4a^2+a+4a^2-4a+1+6a}{4a}\)

= \(\dfrac{a\left(2a-1\right)^2+\left(2a-1\right)^2}{4a}+\dfrac{3}{2}=\dfrac{\left(2a-1\right)^2\left(a+1\right)}{4a}+\dfrac{3}{2}\left(1\right)\)

Vì với a>0 thì\(\dfrac{\left(2a-1\right)^2\left(a+1\right)}{4a}\ge0\)

Dấu = xảy ra khi a=1/2

Nên từ (1) => A\(\ge0+\dfrac{3}{2}\) hay A\(\ge\dfrac{3}{2}\)

Vậy GTNN của A=3/2 khi a=b=1/2

A = \(\dfrac{8a^2+b}{4a}+b^2\)

Ta có: a + b \(\ge\) 1 \(\Leftrightarrow\) b \(\ge\) 1 - a

\(\Rightarrow\) A \(\ge\) \(\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2\)

\(\Leftrightarrow\) A \(\ge\) 2a + \(\dfrac{1}{4a}\) - \(\dfrac{1}{4}\) + 1 - 2a + a²

\(\Leftrightarrow\) A \(\ge\) a² + \(\dfrac{1}{4a}\) + \(\dfrac{3}{4}\)

\(\Leftrightarrow\) A \(\ge\) a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) + \(\dfrac{3}{4}\)

Áp dụng BĐT Cô-si cho 3 số dương a²; \(\dfrac{1}{8a}\); \(\dfrac{1}{8a}\)

a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) \(\ge\) 3\(\sqrt[3]{\dfrac{a^2}{64a^2}}\) = 3\(\sqrt[3]{64}\) = 3.4 = 12

\(\Leftrightarrow\) a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) + \(\dfrac{3}{4}\) \(\ge\) 12 + \(\dfrac{3}{4}\) = \(\dfrac{51}{4}\)

Hay A \(\ge\) a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) + \(\dfrac{3}{4}\) \(\ge\) \(\dfrac{51}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\) a² = \(\dfrac{1}{8a}\) \(\Leftrightarrow\) 8a³ = 1 \(\Leftrightarrow\) a³ = \(\dfrac{1}{8}\) \(\Leftrightarrow\) a = \(\dfrac{1}{2}\)

và b = 1 - a \(\Leftrightarrow\) b = 1 - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\)

Vậy Min_A = \(\dfrac{51}{4}\) \(\Leftrightarrow\) a = b = \(\dfrac{1}{2}\)

 Chúc bn học tốt! (ko chắc lắm đâu)

Với ab = 1 , a + b ¹ 0, ta có:

P = a 3 + b 3 ( a + b ) 3 ( a b ) 3 + 3 ( a 2 + b 2 ) ( a + b ) 4 ( a b ) 2 + 6 ( a + b ) ( a + b ) 5 ( a b ) = a 3 + b 3 ( a + b ) 3 + 3 ( a 2 + b 2 ) ( a + b ) 4 + 6 ( a + b ) ( a + b ) 5 = a 2 + b 2 − 1 ( a + b ) 2 + 3 ( a 2 + b 2 ) ( a + b ) 4 + 6 ( a + b ) 4 = ( a 2 + b 2 − 1 ) ( a + b ) 2 + 3 ( a 2 + b 2 ) + 6 ( a + b ) 4 = ( a 2 + b 2 − 1 ) ( a 2 + b 2 + 2 ) + 3 ( a 2 + b 2 ) + 6 ( a + b ) 4 = ( a 2 + b 2 ) 2 + 4 ( a 2 + b 2 ) + 4 ( a + b ) 4 = ( a 2 + b 2 + 2 ) 2 ( a + b ) 4 = ( a 2 + b 2 + 2 a b ) 2 ( a + b ) 4 = ( a + b ) 2 2 ( a + b ) 4 = 1

Vậy P = 1, với ab = 1 , a+b ¹ 0.