Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta áp dụng công thức \(log_a^{x_1x_2...x_n}=log_a^{x_1}+log_a^{x_2}+...+log_a^{x_n}\) ta có
\(log_2^{600}=log_2^{25.8.3}=log_2^{25}+log_2^8+log_2^3=2log_2^5+3+log_2^3=2b+3+a\)
ta có \(log^{27}_2=log^{3^3}_2=3log^3_2=a\Rightarrow log^3_2=\frac{a}{3}\)
mặt khác
\(log^{\sqrt[6]{2}}_{\sqrt{3}}=\frac{1}{log^{\sqrt{3}}_{\sqrt[6]{2}}}=\frac{1}{log^{3^{\frac{1}{2}}}_{2^{\frac{1}{6}}}}=\frac{1}{\frac{1}{2}log^3_{2^{\frac{1}{6}}}}=\frac{1}{\frac{1}{2}\frac{1}{\frac{1}{6}}log_2^3}=\frac{1}{3.log_2^3}=\frac{1}{3}.\frac{a}{3}=\frac{a}{9}\)
ta có \(\left(log^b_a+log^a_b+2\right)\left(log^b_a-log_{ab}^b\right).log_b^a-1=\left(log^b_a+log^a_b+2\right)\left(log^b_a.log_b^a-log_{ab}^b.log_b^a\right)-1=\left(log^b_a+log^a_b+2\right)\left(1-\frac{1}{log_b^{ba}}log_b^a\right)-1=\left(log^b_a+log^a_b+2\right)\left(1-\frac{1}{1+log^a_b}log^a_b\right)-1=\left(log^b_a+log^a_b+2\right)\frac{1}{1+log^a_b}-1=\left(log^a_b+\frac{1}{log^a_b}+2\right)\frac{1}{1+log^a_b}-1=\frac{\left(1+log^a_b\right)^2}{log^a_b}\frac{1}{1+log^a}-1=\frac{1+log^a_b}{log_b^a}-1=\frac{1}{log_b^a}\)
ta có:
\(\left(log^b_a+\frac{1}{log^b_a}+2\right)\left(log^b_a-\frac{1}{log^{ab}_a}\right)log^a_b-1\)\(=\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(log^b_a-\frac{1}{1+log^b_a}\right)log^a_b-1\)\(=\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(1-\frac{log^a_b}{1+log^b_a}\right)-1\)\(==\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(\frac{1}{1+log^b_a}\right)-1=\frac{1+log^b_a}{log^b_a}-1=\frac{1}{log^b_a}\)
cho 2014=2013+1 thay vào ta có:\(B=x^{2013}-\left(2013+1\right)x^{2012}+\left(2013+1\right)x^{2011}-...-\left(2013+1\right)x^2+\left(2013+1\right)x-1\)
\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...-x^3-x^2+x^2+x-1\)
\(=x-1=2013-1=2012\)
ta có:
\(log^{\left(2a^2\right)}_2+\left(log_2^a\right)a^{log_a^{\left(log^a_1+1\right)}}+\frac{1}{2}log^2_2a^4=log_2^2+log_2^{a^2}+log_2^a\left(log^a_2+1\right)+\frac{1}{2}log^2_2a^4\)
\(=1+2log^a_2+log^a_2\left(1+log^a_2\right)+2log^2a_2\)
\(=3log^2_2a+3log^a_2+1\)
15.
Ta có \(a+b+c+ab+bc+ac=6\)
Mà \(ab+bc+ac\le\left(a+b+c\right)^2\)
=> \(\left(a+b+c\right)^2+\left(a+b+c\right)-6\ge0\)
=> \(a+b+c\ge3\)
\(A=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge a^2+b^2+c^2\ge\frac{1}{3}\left(a+b+c\right)^2\ge3\)(ĐPCM)
Bài 18, Đặt \(\left(a^2-bc;b^2-ca;c^2-ab\right)\rightarrow\left(x;y;z\right)\) thì bđt trở thành
\(x^3+y^3+z^3\ge3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz\ge0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\ge0\)
\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\)
Vì \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)nên ta đi chứng minh \(x+y+z\ge0\)
Thật vậy \(x+y+z=a^2-bc+b^2-ca+c^2-ab\)
\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)(đúng)
Tóm lại bđt được chứng minh
Dấu "=": tại a=b=c
Giải:
Vì \(0\leq a,b,c\leq 1\Rightarrow ab,ac,ab\geq abc\)
Do đó mà \(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\leq \frac{a+b+c}{abc+1}\)
Giờ chỉ cần chỉ ra \(\frac{a+b+c}{abc+1}\leq 2\). Thật vậy:
Do \(0\leq b,c\leq 1\Rightarrow (b-1)(c-1)\geq 0\Leftrightarrow bc+1\geq b+c\Rightarrow bc+a+1\geq a+b+c\)
Suy ra \( \frac{a+b+c}{abc+1}\leq \frac{bc+a+1}{abc+1}=\frac{bc+a-2abc-1}{abc+1}+2=\frac{(bc-1)(1-a)-abc}{abc+1}+2\)
Ta có \(\left\{\begin{matrix}bc\le1\\a\le1\\abc\ge0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}\left(bc-1\right)\left(1-a\right)\le1\\-abc\le0\end{matrix}\right.\) \(\Rightarrow \frac{(bc-1)(1-a)-abc}{abc+1}+2\leq 2\Rightarrow \frac{a+b+c}{abc+1}\leq 2\)
Chứng minh hoàn tất
Dấu bằng xảy ra khi \((a,b,c)=(0,1,1)\) và hoán vị.
Chọn C.
Phương pháp : Đưa về phương trình mũ tìm a, b