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Câu hỏi của Duyen Đao - Toán lớp 9 | Học trực tuyến
Đặt: \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(S=\frac{\frac{1}{x}}{\sqrt{\frac{1}{y}.\frac{1}{z}\left(1+\frac{1}{x^2}\right)}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{z}.\frac{1}{x}\left(1+\frac{1}{y^2}\right)}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{x}.\frac{1}{y}\left(1+\frac{1}{z^2}\right)}}\)
\(=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
\(=\sqrt{\frac{yz}{xy+yz+zx+x^2}}+\sqrt{\frac{zx}{xy+yz+zx+y^2}}+\sqrt{\frac{xy}{xy+yz+zx+z^2}}\)
\(=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\frac{zx}{\left(y+x\right)\left(y+z\right)}}+\sqrt{\frac{xy}{\left(z+x\right)\left(z+y\right)}}\)
\(\le\frac{1}{2}.\left(\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}+\frac{x}{x+y}+\frac{x}{z+x}+\frac{y}{z+y}\right)\)
\(=\frac{1}{2}.\left(1+1+1\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{3}\)
Cần chứng minh: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)\)
Thật vậy: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)^2\Leftrightarrow4\left(a^2-ab+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow4a^2-4ab+4b^2-a^2-b^2-2ab\ge0\Leftrightarrow3\left(a^2+b^2-2ab\right)\ge0\Leftrightarrow3\left(a-b\right)^2\ge0\)(đúng)
Áp dụng:\(P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ac+a^2}}\)
\(\le\frac{1}{\frac{1}{2}\left(a+b\right)}+\frac{1}{\frac{1}{2}\left(b+c\right)}+\frac{1}{\frac{1}{2}\left(c+a\right)}=2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=3\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Ta có:
\(P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ca+a^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(b+c\right)^2+\frac{3}{4}\left(b-c\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(c+a\right)^2+\frac{3}{4}\left(c-a\right)^2}}\)
\(\le2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\le2.\frac{1}{4}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
B3 mk tìm đc cách giải r nhưng bạn nào muốn thì trả lời cg đc
Các bạn giải giúp mình B2 và B5 nhé. Mấy bài kia mình giải được rồi.
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
\(\sqrt{a^2+\dfrac{1}{b+c}}=\dfrac{2}{\sqrt{17}}\sqrt{\left(4+\dfrac{1}{4}\right)\left(a^2+\dfrac{1}{b+c}\right)}\ge\dfrac{2}{\sqrt{17}}\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)\)
\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}\right)\)
\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)\)
Mặt khác:
\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\)
\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{6\left(a+b+c\right)}}\right)\)
\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}\left(a+b+c\right)+\dfrac{a+b+c}{8}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}\right)\)
\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}.6+3\sqrt[3]{\dfrac{81\left(a+b+c\right)}{32.6.\left(a+b+c\right)}}\right)=\dfrac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra khi \(a=b=c=2\)