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1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
1)\(\frac{1}{2\text{a}}=\frac{1.\text{x^2}}{2\text{a.x}^2}=\frac{x^2}{2\text{ax}^2};\frac{2}{x}=\frac{2.2\text{a}x}{x.2\text{ax}}=\frac{4\text{ax}}{2\text{ax}^2}\)\(;\frac{x^2-2\text{ã}}{2\text{ax}^2}\)giữ nguyên
2) \(\frac{x}{a-2}=\frac{x.3\text{a}}{3\text{a}\left(a-2\right)}=\frac{3\text{ax}}{3\text{a}^2-6\text{a}};\frac{2}{3\text{a}}=\frac{2.\left(a-2\right)}{3\text{a}\left(a-2\right)}=\frac{2\text{a}-4}{3\text{a}^2-6\text{a}};\frac{5\text{a}-4}{3\text{a}^2-6\text{a}}\)giữ nguyên
3) \(\frac{x}{10\text{x}-10}=\frac{x.3\text{x}}{\left(10\text{x}-10\right).3\text{x}}=\frac{3\text{x}^2}{30\text{x}^2-30};\frac{1}{3\text{x}-3}=\frac{1.10\text{x}}{10\text{x}.\left(3\text{x}-3\right)}=\)\(\frac{10\text{x}}{30\text{x}^2-30\text{x}};\frac{9\text{x}-10}{30\text{x}^2-30\text{x}}\)giữ
4) \(\frac{1}{1-a}==\frac{-1}{a-1}=\frac{-1.\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-a^2-a-1}{a^3-1};\frac{1}{a^2+a+1}=\frac{1.\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{a-1}{a^3-1};\frac{a^3+2}{a^3-1}\)giữ nguyên
Áp dụng Cauchy, ta có:
\(a^4+b^2\ge2\sqrt{a^4b^2}=2a^2b\)
\(\Rightarrow\frac{1}{a^4+b^2+2ab^2}\le\frac{1}{2a^2b+2ab^2}\)
Tượng tự:
\(\frac{1}{b^4+a^2+2a^2b}\le\frac{1}{2a^2b+2ab^2}\)
\(\Rightarrow A\le\frac{2}{2ab\left(a+b\right)}\)
Lại có: \(\frac{1}{a}+\frac{1}{b}=2\)\(\Leftrightarrow\frac{a+b}{ab}=2\Rightarrow a+b=2ab\)
\(\Rightarrow A\le\frac{2}{\left(a+b\right)^2}\)
Áp dụng Schwarzt: \(2=\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge a+b\ge2\Rightarrow\left(a+b\right)^2\ge4\)
\(\Rightarrow A\le\frac{2}{4}=\frac{1}{2}\)
Dấu = xảy ra khi a=b=1
Áp dụng bđt cosi ta có :
A < = 1/2a^2b+2/ab^2 + 1/2ab^2+2a^2b
= 1/2ab . (1/a+b + 1/a+b) = 1/2ab . 2/a+b = 1/(a+b).(ab)
< = 1/\(\sqrt{ab}.2.ab\) = 1/2\(\sqrt{ab}^3\)
Có : 2 = 1/a + 1/b >= 2\(\sqrt{\frac{1}{ab}}\)
=> \(\sqrt{\frac{1}{ab}}\)< = 1
=> 1/ab < = 1
=> ab > =1
=> A < = 1/2.1 = 1/2
Dấu "=" xảy ra <=> a=b=1
Vậy GTLN của A = 1/2 <=> a=b=1
Tk mk nha