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Lời giải:
Xét biểu thức C
Ta có: \(C=x+\frac{4}{(x-y)(y+1)^2}=x-y+y+\frac{4}{(x-y)(y+1)^2}\)
\(C=(x-y)+\frac{y+1}{2}+\frac{y+1}{2}+\frac{4}{(x-y)(y+1)^2}-1\)
Áp dụng BĐT AM-GM ta có:
\((x-y)+\frac{y+1}{2}+\frac{y+1}{2}+\frac{4}{(x-y)(y+1)^2}\geq 4\sqrt[4]{(x-y).\frac{(y+1)^2}{4}.\frac{4}{(x-y)(y+1)^2}}=4\)
\(\Rightarrow C\geq 4-1=3\Leftrightarrow C_{\min}=3\)
Dấu bằng xảy ra khi \(x=2; y=1\)
Biểu thức D không có điều kiện gì thì không có min em nhé. Trừ khi \(D=x+\frac{1}{xy(x-y)}\)
Áp dụng bất đẳng thức AM - GM, ta có:
\(S=\dfrac{1}{\left(x-1\right)^2}+\dfrac{1}{\left(2-x\right)^2}+\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\ge3\sqrt[3]{\dfrac{1}{\left(x-1\right)^2}\times\dfrac{1}{\left(2-x\right)^2}\times\dfrac{1}{\left(x-1\right)\left(2-x\right)}}\)
\(=\dfrac{3}{\left(x-1\right)\left(x-2\right)}=\dfrac{3}{-x^2+3x-2}\)
Vì \(-x^2+3x-2=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
nên \(S\ge\dfrac{3}{\dfrac{1}{4}}=12\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{\left(x-1\right)^2}=\dfrac{1}{\left(2-x\right)^2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\\x-\dfrac{3}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3}{2}\left(\text{ nhận }\right)\)
Vậy \(Min_S=12\Leftrightarrow x=\dfrac{3}{2}\)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
Ta có \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}=\sqrt{\dfrac{2\sqrt{3}+2}{\left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right)}-\dfrac{3\left(2\sqrt{3}-2\right)}{\left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2\left(\sqrt{3}+1\right)}{12-4}-\dfrac{2\left(3\sqrt{3}-3\right)}{12-4}}=\sqrt{\dfrac{\sqrt{3}+1}{4}-\dfrac{3\sqrt{3}-3}{4}}=\sqrt{\dfrac{\sqrt{3}+1-3\sqrt{3}+3}{4}}=\sqrt{\dfrac{4-2\sqrt{3}}{4}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{4}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{2}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=\dfrac{\left|\sqrt{3}-1\right|}{2}=\dfrac{\sqrt{3}-1}{2}\Leftrightarrow2x=\sqrt{3}-1\Leftrightarrow2x+1=\sqrt{3}\Leftrightarrow\left(2x+1\right)^2=3\Leftrightarrow4x^2+4x-2=0\Leftrightarrow2x^2+2x-1=0\)
Ta lại có \(P=\dfrac{4\left(x+1\right)x^{2018}-2x^{2017}+2x+1}{2x^2+3x}=\dfrac{2x^{2017}\left[2\left(x+1\right)x-1\right]+\sqrt{3}}{2x^2+2x-1+x+1}=\dfrac{2x^{2017}\left[2x^2+2x-1\right]+\sqrt{3}}{x+1}=\dfrac{\sqrt{3}}{x+1}=\sqrt{3}:\left(x+1\right)=\sqrt{3}:\left(\dfrac{\sqrt{3}-1}{2}+1\right)=\sqrt{3}:\dfrac{\sqrt{3}+1}{2}=\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{2\left(3-\sqrt{3}\right)}{2}=3-\sqrt{3}\)Vậy khi \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}\) thì P=\(3-\sqrt{3}\)