ừ bài nâng cao mà bạn ơi :)))

\(P=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^4+...+\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\)

\(=\left(\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2\right)+\left(\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{4}\right)^4\right)+...+\left(\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\right)\)

\(=\dfrac{1}{3}.\dfrac{2}{3}+\left(\dfrac{1}{3}\right)^3.\dfrac{2}{3}+...+\left(\dfrac{1}{3}\right)^{19}.\dfrac{2}{3}\)

\(=\dfrac{2}{3}.\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{19}\right]\)

B = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{19}{20}\)

= \(\dfrac{1}{20}\)

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

Bạn ơi mik ra \(\dfrac{x^3+45x-54}{12\left(x-3\right)\left(x+3\right)}\) có đúng không bạn?

Mình rút chx hết bạn bạn gửi cách làm bạn qua mình tham khảo đc k ạ?

Sửa đề: \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(a,C=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\left(a>0;a\ne1;a\ne4\right)\\ C=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,C\ge\dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}\ge0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}\ge0\\ \Leftrightarrow\sqrt{a}-4\ge0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a\ge16\)

Đặt \(A=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)...\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)...\left(20^4+\dfrac{1}{4}\right)}\)

\(=\dfrac{\left[\left(1^4+\dfrac{1}{4}\right).2^4\right]\left[\left(3^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(19^4+\dfrac{1}{4}\right).2^4\right]}{\left[\left(2^4+\dfrac{1}{4}\right).2^4\right]\left[\left(4^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(20^4+\dfrac{1}{4}\right).2^4\right]}\)

\(=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)

Lưu ý: \(a^4+4=\left(a^4+4a^2+4\right)-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2\)

\(=\left(a^2-2a+2\right)\left(a^2+2a+2\right)\)

Áp dụng vào biểu thức A, ta có:

\(A=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)

\(=\dfrac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)...\left(38^2-38.2+2\right)\left(38^2+38.2+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)...\left(40^2-2.40+2\right)\left(40^2+2.40+2\right)}\)

\(=\dfrac{2.10.26..1370.1522}{10.26.50...1522.1682}=\dfrac{2}{1682}=\dfrac{1}{841}\)

Vậy \(A=\dfrac{1}{841}\)

mk thấy cũng khá đơn giản mà Phạm Ngọc Diễm

\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)

a) \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(a>0.a\ne1\right)\)

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1-\sqrt{a}-2}{\sqrt{a}-1}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(1-\sqrt{a}\right)=-\dfrac{1}{\sqrt{a}}\)

b) \(C=\dfrac{1}{4}\Rightarrow-\dfrac{1}{\sqrt{a}}=\dfrac{1}{4}\Rightarrow\sqrt{a}=-4\) (vô lý) \(\Rightarrow\) không có a thỏa đề