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I think that we have to prove \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)
We have \(a+b+c=abc\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
We have \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=0\)( Because \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\))
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)
So...
Em chỉ giải ra được 1 TH dấu bằng thôi: a = b = c (còn trường hợp a = b; c=0 và các hoán vị thì em chịu, vì khi xét dấu = trong bđt thì em chỉ xảy ra 1 th)
Áp dụng BĐT Cauchy-Schwarz dạng Engel;
\(VT\ge\frac{16}{a^2+b^2+c^2+\left(a+b+c\right)^2}\ge\frac{16}{\frac{\left(a+b+c\right)^2}{3}+\left(a+b+c\right)^2}\)\(=\frac{12}{\left(a+b+c\right)^2}\) (đpcm)
Đẳng thức xảy ra khi a = b = c
a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)
b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)
c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)
\(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)
d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)
14.
\(log_aa^2b^4=log_aa^2+log_ab^4=2+4log_ab=2+4p\)
15.
\(\frac{1}{2}log_ab+\frac{1}{2}log_ba=1\)
\(\Leftrightarrow log_ab+\frac{1}{log_ab}=2\)
\(\Leftrightarrow log_a^2b-2log_ab+1=0\)
\(\Leftrightarrow\left(log_ab-1\right)^2=0\)
\(\Rightarrow log_ab=1\Rightarrow a=b\)
16.
\(2^a=3\Rightarrow log_32^a=1\Rightarrow log_32=\frac{1}{a}\)
\(log_3\sqrt[3]{16}=log_32^{\frac{4}{3}}=\frac{4}{3}log_32=\frac{4}{3a}\)
11.
\(\Leftrightarrow1>\left(2+\sqrt{3}\right)^x\left(2+\sqrt{3}\right)^{x+2}\)
\(\Leftrightarrow\left(2+\sqrt{3}\right)^{2x+2}< 1\)
\(\Leftrightarrow2x+2< 0\Rightarrow x< -1\)
\(\Rightarrow\) có \(-2+2020+1=2019\) nghiệm
12.
\(\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\0< log_3\left(x-2\right)< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\1< x-2< 3\end{matrix}\right.\)
\(\Rightarrow3< x< 5\Rightarrow b-a=2\)
13.
\(4^x=t>0\Rightarrow t^2-5t+4\ge0\)
\(\Rightarrow\left[{}\begin{matrix}t\le1\\t\ge4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4^x\le1\\4^x\ge4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le0\\x\ge1\end{matrix}\right.\)
\(\frac{a^2}{b-1}+4\left(b-1\right)\ge4a\) ; \(\frac{b^2}{c-1}+4\left(c-1\right)\ge4b\); \(\frac{c^2}{a-1}+4\left(a-1\right)\ge4c\)
Cộng vế với vế, chuyển vế và rút gọn ta có đpcm
Dấu "=" xảy ra khi \(a=b=c=2\)