Đặt \(a+\dfrac{1}{36a}=x\)

pt đã cho trở thành 9x² - 6x + 1 = 0

\(\Leftrightarrow9\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow9\left(x-\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{3}=0\)

\(\Leftrightarrow x=\dfrac{1}{3}=a+\dfrac{1}{36a}=\dfrac{36a^2+1}{36a}\)

\(\Leftrightarrow12a=36a^2+1\)

\(\Leftrightarrow36a^2-12a+1=0\)

\(\Leftrightarrow\left(6a-1\right)^2=0\)

\(\Leftrightarrow6a-1=0\)

\(\Leftrightarrow a=\dfrac{1}{6}\) \(\Rightarrow a=6\)

\(\frac{1}{x+1}+\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}\) (ĐKXĐ: \(x\ne1;-1\))

\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x-1\right)\left(x^2-1\right)}=-\frac{3}{x^2-1}\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\frac{2}{\left(x-1\right)^2\left(x+1\right)}=-\frac{3}{x^2-1}\)

\(\Leftrightarrow\frac{x^2+1-2x+2}{\left(x-1\right)^2\left(x+1\right)}=-\frac{3}{x^2-1}\)

\(\Leftrightarrow\frac{x^2-2x+3}{x-1}=-3\)

\(\Leftrightarrow x^2-2x+3=-3x+3\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Rightarrow\left[\begin{matrix}x=0\left(TM\right)\\x+1=0\left(KTM\right)\end{matrix}\right.\)

Vậy ...

a) \(\frac{1.3.5...39}{21.22.23...40}=\frac{1.2.3.4.5.6...39.40}{\left(2.4.6...40\right).21.22.23...40}=\frac{1.2.3.4.5.6...39.40}{2^{20}.1.2.3...20.21.22.23...40}\)

\(=\frac{1}{2^{20}}\left(đpcm\right)\)

b) \(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{\left(2.4.6...2n\right)\left(n+1\right)\left(n+2\right)...2n}=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{2^n.1.2.3...n\left(n+1\right)\left(n+2\right)...2n}\)

\(=\frac{1}{2^n}\left(đpcm\right)\)

Dạng tổng quát: \(\sqrt[k+1]{\frac{k+1}{k}}>\sqrt[k+1]{\frac{k+1}{k+1}}=1\) với k = 1; 2; 3; ...; n

=> \(a=\sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+...+\sqrt[n+1]{\frac{n+1}{n}}>n\) (1)

Áp dụng bđt AM-GM cho k + 1 số dương ta có:

\(\sqrt[k+1]{\frac{k+1}{k}}=\sqrt[k+1]{1.1.1...1.\frac{k+1}{k}}< \frac{1+1+1+...+1+\frac{k+1}{k}}{k+1}=\frac{1.k}{k+1}+\frac{\frac{k+1}{k}}{k+1}\)

\(\Leftrightarrow\sqrt[k+1]{\frac{k+1}{k}}< \frac{k}{k+1}+\frac{1}{k}=1-\frac{1}{k+1}+\frac{1}{k}=1+\left(\frac{1}{k}-\frac{1}{k+1}\right)\)

\(< 1+\frac{1}{k\left(k+1\right)}\)

Áp dụng vào bài ta được:

\(a< \left(1+\frac{1}{1.2}\right)+\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+...+\left(1+\frac{1}{n\left(n+1\right)}\right)\)

\(a< n+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)\)

\(a< n+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)

\(a< n+\left(1-\frac{1}{n+1}\right)< n+1\) (2)

Từ (1) và (2) suy ra phần nguyên của a là n

Ôi cao thủ, đề ko cho a;b >= 0, lm cx ko cho thêm đk vô mà vx lm như đúng rồi ấy

Lấy đơn giản a = b = 1; c = -2 thì sao ???

༺ ๖ۣۜPhạm ✌Tuấn ✌Kiệτ ༻: đề cần cho thêm a;b;c không âm

Tìm Max

Chia cả 2 vế của Q cho b² ta có:

\(Q=\dfrac{\left(\dfrac{a}{b}\right)^2-\dfrac{a}{b}+1}{\left(\dfrac{a}{b}\right)^2+\dfrac{a}{b}+1}\)

Đặt \(x=\dfrac{a}{b}\), lúc này ta có: \(Q=\dfrac{x^2-x+1}{x^2+x+1}\)\(=\dfrac{3x^2+3x+3-2x^2-4x-2}{x^2+x+1}=3-\dfrac{2\left(x+1\right)^2}{x^2+x+1}\le3\)

Dấu "=" xảy ra khi x = -1 = \(\dfrac{a}{b}\Leftrightarrow a=-b\)

Tìm Min

\(Q=1-\dfrac{2ab}{a^2+ab+b^2}\ge1-\dfrac{2ab}{2ab+ab}=1-\dfrac{2}{3}=\dfrac{1}{3}\)

Dấu "=" xảy ra khi a = b

Áp dụng bđt Cauchy-Schwraz dạng Engel ta có:

\(P=\left(1-\dfrac{1}{x+1}\right)+\left(1-\dfrac{1}{y+1}\right)+\left(1-\dfrac{1}{z+1}\right)\)

\(=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\le3-\dfrac{\left(1+1+1\right)^2}{x+1+y+1+z+1}\)

\(\le3-\dfrac{9}{4}=\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

b) x² + y² + z² = xy + xz + yz

<=> 2x² + 2y² + 2z² - 2xy - 2xz - 2yz = 0

<=> (x² + y² - 2xy) + (y² + z² - 2yz) + (x² + z² - 2xz) = 0

<=> (x - y)² + (y - z)² + (x - z)² = 0

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(x-z\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\)\(\Leftrightarrow x=y=z\)

Theo đề bài lại có: x + y + z = 2013

Do đó, \(3x=3y=3z=2013\)\(\Rightarrow x=y=z=671\)

2) a) \(\frac{x^2-5x+1}{2x+1}+2=-\frac{x^2-4x+1}{x+1}\) (ĐKXĐ: \(x\ne-\frac{1}{2};-1\))

+) x = \(-\frac{2}{3}\), thay vào đề không TM

+ x\(\ne-\frac{2}{3}\)

Từ đề \(\Rightarrow\frac{x^2-5x+1+4x+2}{2x+1}=\frac{-x^2+4x-1}{x+1}\)

\(\Leftrightarrow\frac{x^2-x+3}{2x+1}=\frac{-x^2+4x-1}{x+1}=\frac{\left(x^2-x+3\right)+\left(-x^2+4x-1\right)}{\left(2x+1\right)+\left(x+1\right)}\) \(=\frac{3x+2}{3x+2}=1\)

\(\Rightarrow x^2-x+3=2x+1\)

\(\Leftrightarrow x^2-3x+2=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{1}{4}\)

\(\Rightarrow\left[\begin{matrix}x-\frac{3}{2}=\frac{1}{2}\\x-\frac{3}{2}=-\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=2\\x=1\end{matrix}\right.\)

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