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điện thoại cùi nên chụp hơi mờ, đề này còn thiếu a,,bc>0
\(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(P=\sqrt{\dfrac{yz}{xy+xz}}+\sqrt{\dfrac{zx}{xy+yz}}+\sqrt{\dfrac{xy}{yz+zx}}\)
\(P=\dfrac{2yz}{2\sqrt{yz\left(xy+xz\right)}}+\dfrac{2zx}{2\sqrt{zx\left(xy+yz\right)}}+\dfrac{2xy}{2\sqrt{xy\left(yz+zx\right)}}\)
\(P\ge\dfrac{2yz}{xy+yz+zx}+\dfrac{2zx}{xy+yz+zx}+\dfrac{2xy}{xy+yz+zx}=2\)
Dấu "=" không xảy ra nên \(P>2\)
Từ \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2017\)
\(\Leftrightarrow7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2017\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le2017\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(T=\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2c^2+a^2\right)}}\)
\(\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2^2}{2a}+\dfrac{1^2}{b}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2b}+\dfrac{1^2}{c}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2c}+\dfrac{1^2}{a}\right)\)
\(\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\)\(=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\le\sqrt{\left(\dfrac{1}{81}+\dfrac{1}{81}+\dfrac{1}{81}\right)\left(\dfrac{9}{a^2}+\dfrac{9}{b^2}+\dfrac{9}{c^2}\right)}\)
\(\le\sqrt{\dfrac{1}{81}\cdot3\cdot9\cdot2017}=\sqrt{\dfrac{2017}{3}}\)
Vậy \(T_{Max}=\sqrt{\dfrac{2017}{3}}\) khi \(a=b=c=\sqrt{\dfrac{3}{2017}}\)
So kimochiii~
a) Ta có: \(A=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^2-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^4-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^8-1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{16}-1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{32}-1\right]\)
\(=65535\sqrt{2}+65535\)
b) Ta có: \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2020}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)
\(=\sqrt{2020}-1\)
\(=2\sqrt{505}-1\)
c) Ta có: \(C^3=26+15\sqrt{3}+26-15\sqrt{3}+3\cdot\sqrt[3]{\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\cdot\left(\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\right)\)
\(\Leftrightarrow C^3=52+3\cdot C\)
\(\Leftrightarrow C^3-3\cdot C-52=0\)
\(\Leftrightarrow C^3-4C^2+4C^2-16C+13C-52=0\)
\(\Leftrightarrow C^2\left(C-4\right)+4C\left(C-4\right)+13\left(C-4\right)=0\)
\(\Leftrightarrow\left(C-4\right)\left(C^2+4C+13\right)=0\)
mà \(C^2+4C+13>0\)
nên C-4=0
hay C=4
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)}\)
Đặt BT đề cho là P
\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
Áp dụng BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(\text{luôn đúng}\right)\)
Ta có \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Cộng VTV \(\Leftrightarrow3\ge\dfrac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\sqrt[3]{abc}+1\)
\(\Leftrightarrow VT^2=\sum\left[\dfrac{1}{a\left(b+1\right)}\right]^2\ge3\cdot\sum\dfrac{1}{ab\left(a+1\right)\left(b+1\right)}\\ \Leftrightarrow VT^2\ge3\cdot\dfrac{a^2+b^2+c^2+a+b+c}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge3\cdot\dfrac{a+b+c+ab+bc+ca}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\\ \Leftrightarrow VT^2\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(1+\sqrt[3]{abc}\right)^3}\\ \Leftrightarrow VT^2\ge\dfrac{9}{\sqrt[3]{\left(abc\right)^2}\left(1+\sqrt[3]{abc}\right)^2}=VP^2\\ \LeftrightarrowĐpcm\)
Dấu \("="\Leftrightarrow a=b=c=1\)