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\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)
\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)
\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)
\(=\frac{1}{x}-\frac{1}{x-5}\)
\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)
\(=\frac{x-5-x}{x\left(x-5\right)}\)
\(=-\frac{5}{x\left(x-5\right)}\)
a) Vậy A = 3/4 <=> x = -1/2 A = x 2 + x + 1 A = x 2 + 2. x + + 1 2 1 4 3 4 A = (x + ) 2 + ≥ 1 2 3 4 3 4
\(A=x^2-x+1\)
\(=\left(x^2-2x\frac{1}{2}+\frac{1}{4}\right)+1-\frac{1}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu"=" xảy ra khi \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Vậy \(Min_A=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(P=\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}+\frac{1}{1-x}\)
ĐKXĐ : \(x\ne1\)
\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\)
\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x}{x^2+x+1}\)
b) Xét hiệu P - 1/3 ta có :
\(\frac{x}{x^2+x+1}-\frac{1}{3}=\frac{3x}{3\left(x^2+x+1\right)}-\frac{x^2+x+1}{3\left(x^2+x+1\right)}=\frac{3x-x^2-x-1}{3\left(x^2+x+1\right)}=\frac{-x^2+2x-1}{3\left(x^2+x+1\right)}\)
\(=\frac{-\left(x^2-2x+1\right)}{3\left(x^2+x+1\right)}=\frac{-\left(x-1\right)^2}{3\left(x^2+x+1\right)}\)
Ta có : ( x - 1 )2 ≥ 0 ∀ x => -( x - 1 )2 ≤ 0 ∀ x
x2 + x + 1 = ( x2 + x + 1/4 ) + 3/4 = ( x + 1/2 )2 + 3/4 ≥ 3/4 > 0 ∀ x
=> 3( x2 + x + 1 ) ≥ 9/4 > 0 ∀ x
Vậy -( x - 1 )2 và 3( x2 + x + 1 ) trái dấu nhau
=> \(\frac{-\left(x-1\right)^2}{3\left(x^2+x+1\right)}\le0\)hay P - 1/3 ≤ 0
Đẳng thức xảy ra <=> x = 1 ( ktm ) => Không xảy ra đẳng thức
Vậy P < 1/3 ( đpcm )
\(P=\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}+\frac{1}{1-x}\)
\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x}{x^2+x+1}\)
a) \(\frac{x-1}{x^2-1}-\frac{x+1}{x^2+x}=\frac{x-1}{\left(x-1\right)\left(x+1\right)}-\frac{x+1}{x\left(x+1\right)}\)
\(=\frac{1}{x+1}-\frac{1}{x}\)
\(=\frac{x}{x\left(x+1\right)}-\frac{x+1}{x\left(x+1\right)}\)
\(=\frac{x-x-1}{x\left(x+1\right)}=\frac{-1}{x\left(x+1\right)}\)
b) \(\frac{2x+2y}{y-x}-\frac{x^2+xy}{3x^2-3y^2}=\frac{-2x-2y}{x-y}-\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}\)
\(=\frac{-2x-2y}{x-y}-\frac{x\left(x+y\right)}{3\left(x-y\right)\left(x+y\right)}\)
\(=\frac{-2x-2y}{x-y}-\frac{x}{3\left(x-y\right)}\)
\(=\frac{3\left(-2x-2y\right)}{3\left(x-y\right)}-\frac{x}{3\left(x-y\right)}\)
\(=\frac{-6x-6y}{3\left(x-y\right)}-\frac{x}{3\left(x-y\right)}\)
\(=\frac{-7x-6y}{3\left(x-y\right)}\)
a, \(\frac{x-1}{x^2-1}-\frac{x+1}{x^2+x}=\frac{x-1}{\left(x-1\right)\left(x+1\right)}-\frac{x+1}{x\left(x+1\right)}\)
\(=\frac{x\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2-x-x^2+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}=\frac{-1}{x\left(x+1\right)}\)
b, \(\frac{2x+2y}{y-x}-\frac{x^2+xy}{3x^3-3y^2}=-\frac{2x+2y}{x-y}-\frac{x^2+xy}{3x\left(x^2-y^2\right)}\)
\(=-\frac{2x+2y}{x-y}-\frac{x^2+xy}{3x\left(x-y\right)\left(x+y\right)}\)
\(=-\frac{6x\left(x+y\right)^2}{3x\left(x-y\right)\left(x+y\right)}-\frac{x^2+xy}{3x\left(x-y\right)\left(x+y\right)}\)
\(=-\frac{6x\left(x^2+2xy+y^2\right)}{3x\left(x-y\right)\left(x+y\right)}-\frac{x^2+xy}{3x\left(x-y\right)\left(x+y\right)}\)
\(=\frac{-12x^3-12x^2y-6xy^2-x^2-xy}{3x\left(x-y\right)\left(x+y\right)}\)
check hộ ý b nhá :))
Ta có:\(a^{2000}+b^{2000}=a^{2001}+b^{2001}=a^{2002}+b^{2002}\)
\(\Rightarrow a^{2000}+b^{2000}+a^{2002}+b^{2002}=2\left(a^{2001}+b^{2001}\right)\)
\(\Rightarrow a^{2002}-a^{2001}-a^{2001}+a^{2000}+b^{2002}-b^{2001}-b^{2001}+b^{2000}=0\)
\(\Rightarrow a^{2001}\left(a-1\right)-a^{2000}\left(a-1\right)+b^{2001}\left(b-1\right)-b^{2000}\left(b-1\right)=0\)
\(\Rightarrow\left(a-1\right)\left(a^{2001}-a^{2000}\right)+\left(b-1\right)\left(b^{2001}-b^{2000}\right)=0\)
\(\Rightarrow a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2=0\)
Dấu"="xảy ra khi \(\orbr{\begin{cases}a^{2000}\left(a-1\right)^2=0\\b^{2000}\left(b-1\right)^2=0\end{cases}}\)Mà \(a,b>0\)
\(\Rightarrow a=b=1\)
Do đó:\(a^{2020}+b^{2020}=1^{2020}+1^{2020}=1+1=2\)
